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 So far we have discussed calculus of functions of one variable . In this section we deal with functions of two or more variables . However in real life a function depends upon many dependent variables . Such functions , in general are called multi - variable functions or functions of several variables . In this section , we primarily deal with functions of two variables . All the concepts can be analogously extended to functions of two or more variables . Many functions in real life depend upon more than one variables . In chapters 3 and 5 , we discussed how to obtain the derivatives and integrals of functions with one independent variable . In this section , we will consider functions with two or more independent variables . 
 Let us look at some examples : ( i ) The area of a rectangle depends upon the length and width of the rectangle . Therefore , we can think of area as function of two variables , . ( ii ) The temperature of particular place on earth depends on the longitude and the latitude of that point . So we can think of temperature as a function of two variables , say . ( iii ) The cost of a mobile hand - set depends upon many factors . 
 In this section we deal with real valued functions of two real variables . The domain of such functions is a set of points in . We will assume that is endowed with Euclidean distance , that is distance between any two points , and is given by 
 A point in can also be referred to as a vector , which is the position vector of that point . 
 A real valued function of two real variables in general can be defined as follows : 
 Let . A function of two variables is a rule that assigns to each point a unique real number denoted by . The set is called the domain of and its range is the set of values that takes . 
 Generally , we write to make explicit the value taken on by at the point . Here and are called independent variables and is called the dependent variable . 
 Most often the domain of the function is not mentioned explicitly , however it can be obtained from the function ’ s expression by simply looking at set of points on which the function is well defined . Let us look at some examples . 
 Find the domain of the following function : 
 Solution . ( a ) Note that the right handside of the expression works for all values except when the denominator is zero . That is , the domain of this function is 
 ( b ) The domain of this function is set of points such that . That is , . Thus the domain of is . 
 ( c ) The domain of is a set of points such that . See Figure 1 . That is 
 ( d ) We know that is defined only for positive reals . That is . This happens when 
 ( i ) , or ( ii ) , . For ( i ) , . 
 For ( ii ) , . 
 Then the domain of is . Thus . See Figure 2 . 
 ( e ) The domain of this function is when right handside make sense . That when and . The domain is plotted in Figure 3 . 
 Graph of a Multivariate Function 
 Let . Then the graph of is a set of points 
 Usually , the graph of a function is called a surface in . ( See Figure 4 . ) In Figure 5 the surface is the graph of the polynomial function is plotted . 
 Figure 4 Graph of a function of and is a surface Figure 5 Graph of a polynomial function 
 Unlike one variable function , plotting the graph of a function in is not an easy task . One way to plot is to look at set of points for various values of , which is the intersection of the surface and the plane . The intersection is called the level curve at the level . If we know these level curves for all values of , then we can visualize the surface . 
 Note that level curve is the set of all points in the domain of at which takes value . Level curves are also called contour lines . Thus to plot the graph of a surface all we need is to plot level curves ( contour lines ) for various values of and then put these level curves at the corresponding level planes . ( See Figure 6 ( a ) . ) 
 A contour map of a function is a 2 - dimensional graph showing several level curves corresponding to several values of . A contour map of lies in the - plane . ( See Figure 6 ( b ) . ) 
 Figure 6 Surface in ( a ) and level curves in ( b ) 
 Plot the surface and level curves of . 
 Figure 7 Surface and level curves in Example 2 
 In most instances the task of graphing level curves of a function of two variables is formidable . A CAS was used to generate the surfaces and corresponding level curves in Figure 8 . 
 The level curves of a function are also called contour lines . On a practical level , contour maps are often used to display curves of equal elevation . In Figure 9 , we see that a contour map illustrates the various segments of a hill that have a given altitude . This shows heights of the mountain in in Figure 10 , which show the volcano Mt . Halla , in the state of Jeju , Korea . 
 Figure 9 Contour map of Mt . Halla Figure 10 Map of Mt . Halla in the state of Jeju , South Korea 
 CAS . Graph the level curves of the following functions along with the surface : 
 ( 1 ) http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 1 - Exm - 3 . html 
 ( 3 ) [ CAS ] http : / / sage . skku . edu / 또는 https : / / sagecell . sagemath . org / 
 ( 4 ) Level curves along with the surface : 
 Graph and plot the level curves , and in the domain of in the plane . 
 Solution . The domain of is the entire - plane , and the range of is the set of real numbers less than or equal to . The graph is the paraboloid , a portion of which is shown in Figure 15 . In Figure 16 , The level curve is the set of points in the plane such that . 
 This is the circle of radius centered at the origin . Similarly , the level curves and are the circles 
 The level curve consists of the origin alone . ■ 
 The definition of functions of three or more variables are simply generalizations of the definition of one variable . For example , a function of three variables is a rule of correspondence that assigns to each ordered triple of real number in a subset of , one and only one number in the set of real numbers . A function of three variables is usually denoted by or . 
 For example , the volume and surface area S of a rectangular box are polynomial functions of three variables : 
 The set of points in space where a function of three independent variables has a value is called a level surface of . 
 Describe the level surfaces of the function . 
 Solution . The value of is the distance from the origin to the point . Each level surface is a half - sphere of radius centered at the origin . The level surface consists of the origin alone . ■ 
 ( a ) The level surfaces of the polynomial are a family of parallel planes defined by . ( See Figure 18 . ) 
 Figure 18 Level surfaces in ( a ) of Example 6 
 ( b ) The level surfaces of the polynomial are a family of concentric spheres defined by , . See Figure 17 . 
 Figure 17 Level surfaces in ( b ) of Example 6 
 ( c ) The level surfaces of the rational function are given by or . A few members of this family of paraboloids are 
 given in Figure 19 . 
 Figure 19 Level surfaces in ( c ) of Example 6 
 ( a ) http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 1 - Exm - 6 . html 
 for k , col in [ ( 1 , ' red '),( 2 , ' orange '),( 3 , ' yellow '),( 4 , ' green ')]: 
 p # Figure 17 
 13 . 1 EXERCISES ( Multivariate Functions ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 1 - Sol . html http : / / youtu . be / As _ 0 AYApHlM 
 1 . If , find ( a ) , ( b ) , ( c ) . 
 2 . , find 
 ( a ) , ( b ) . 
 3 - 4 . Find the domain of the given function . 
 5 - 6 . Find the range of the given function . 
 Then the range is . 
 7 - 8 . Sketch typical level surfaces of the function . 
 for k , col in [ ( 1 , ' red '), ( 2 , ' orange '), ( 3 , ' yellow '), ( 4 , ' green ')]: 
 find the , and intercepts of the level surface that passes through . 
 Before discussing limits and continuity , we need to introduce some terminology about sets that will be useful in the sections and chapters that follow . The set 
 is called an open disk centered at with radius . On the other hand , the set 
 A region in is bounded if it can be contained in a sufficiently large disk . 
 Figure 1 Various regions in 
 Let be a function defined on a region . 
 Let and ( The point need not necessarily be in ) . Then we say that the limit of as approaches to is , if for every , there exists such that such that when for . 
 This can be also expressed as 
 Let . Let us check if exists . 
 Solution . It is straightforward to guess that the limit exists and it is . Let be given . To prove that , we need to find such that when , we have . Note that for all . Hence any choice of will work . Thus 
 Solution . It is not hard to guess that when , . Therefore , the limit must be . Let us prove this . Let . We need to find such that when , 
 For any , it is easy to that ( See Figure 2 . ) 
 and . Thus we have 
 ( Triangle Law of Modulus ) 
 This suggests that we may choose . Now if 
 , from we have . Hence , exists and it is . ■ 
 ( 1 ) Let . Then show that . 
 ( 2 ) Let . Then show that . 
 Solution . ( 1 ) Let . We need to find such that when 
 For any , it is easy to that 
 , from we have 
 ( 2 ) Similar to ( 1 ) , we can prove . ■ 
 The notion of “ approaching ” a point is not as simple as it is for functions of one variable . In the - plane , there are an infinite number of ways of approaching a point , as shown in Figure 3 . In order that exist , we require that approach to along every possible curve or path through . In particular , 
 ( ⅰ ) If does not approach the same number for two different paths to , then does not exist . 
 ( ⅱ ) If the limit of does not exists along some path through , then limit at does not exist . 
 In the discussion of that follows we shall assume that the function is defined at every point in an open disk centered at but not necessarily at itself . 
 Investigate the limit of 
 Solution . Let and in such a way that . Then along this line 
 Since the limit of the function depends on the manner in which approach to , the function does not have a limit at . For example , for , and for , . ■ 
 Find the limit , if it exists , of the following function as . 
 Solution . ( a ) Since and , 
 Note that iff and . 
 Thus , as , and so 
 ( b ) Let and in such a way that . Then along this line 
 Since the limit of the function depends on the manner in which approach to , the function does not have limit at . For example , for , and for , . ■ 
 Finding the limit of a function directly from the definition is not an easy task . The following are properties of limit of two variable functions which are similar to properties of one variable functions . These properties give a tool to compute limit of various functions . 
 THEOREM 2 The Power Rule 
 Let and be real numbers and suppose that , . Then 
 5 . , if 
 7 . If and are integers , and have no common factors and , then 
 where we have if is even and and are integers . 
 Using the laws of limit in Theorem 2 , find the limit of 
 ( 1 ) at . 
 Solution . ( 1 ) From Theorem 2 , we have we have 
 ( 2 ) From Theorem 2 , we have 
 Find the limit of the following function as . 
 Solution . ( a ) Let . Then one can show that if and only if . We have . 
 ( b ) When the point is on the straight line , . Therefore , the limit of is as approaches along line . This limit changes as changes . Hence , does not have a limit as . ■ 
 Find the limit of the following functions as . 
 Solution . ( a ) Since , , we have 
 Note that iff and 
 Thus , as , and so . 
 ( b ) Letting along any non - vertical line through the origin . Let along the , we have 
 Therefore , depending on the value of , different paths will lead to different values of the limit as . Hence the limit does not exist as . ■ 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 2 - Exm - 9 . html 
 Answer : ( x , y ) |--> - 3 / 7 
 Answer : - 3 / 7 ■ 
 For a function in neighborhood of point , suppose that 
 exist . Let 
 Then and converges to . Furthermore , 
 and can be distinguished by denoting the former as the double limit while the latter as the successive limit . Since is practically a limit of single variable function , we just repeats the limit process twice . 
 Proof . Since , for any given , there exists a number such that if then . 
 With the same argument , since , for any given , there exists a number such that 
 Since , for any given , there exists a number and such that 
 Thus , if , then and simultaneously if , then 
 Likewise , this may be proven in a similar manner for limits taken in varying orders , that is , first , then and first , then . ■ 
 In Example 7 ( b ) and . Thus and exist but two limits are not equal . Then does not exists . 
 The function is one of examples where the converse of Theorem 3 is false . For this function , 
 However , when , respectively . Thus in any domain near , no matter how small , there are points that are assigned and points that are assigned by the function . This means that the limit of the function at does not exists . The may not exist , even if is satisfied . 
 Using Sage , find the limit of 
 ( 3 ) Find the example of . 
 ( 1 ) http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 2 - Exm - 10 . html 
 y ^ 2 + 4 * y + 3 
 -( 4 * y - 1 ) / ( 2 * y + 3 ) 
 ( 3 ) Let . We have 
 Then since , a function is an example of ( 3 ) . ■ 
 The following definition of continuity for function of two variables is analogous to the definition of continuity for a one variable function . 
 Let and . Then is said to be continuous at if . That is , exists and it is equal to the value of the function . If is continuous at every point in , then we say that is continuous on . 
 It is easy to see that functions defined in Examples 1 , 2 and 3 are continuous on . The proof is similar to the solutions given in Examples 1 , 2 and 3 . ■ 
 Determine the points at which the following function is continuous . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 2 - Exm - 12 . html 
 Solution . The function is a rational function , so it is continuous at all points of its domain , which consists of all points of except . In order for to be continuous at , we must show that 
 You can verify that as approaches along paths of the form , where is any constant , the function values approach . Now consider parabolic paths of the form , where is a nonzero constant . ( See Figure 6 . ) 
 For the choice of , notice that if is replaced by in , the result involves the same power of ( in this case , ) in the numerator and denominator were canceled . 
 We see that along parabolic paths , the limit depends on the approach path . For example , with , along the path , the function values approach . ( See Figure 7 . ) Because function values approach two different numbers along two different paths , the limit at does not exist , and is not continuous at . ■ 
 Investigate the continuity of 
 Hence it is continuous at . ■ 
 Using the properties of limits in Theorem 2 , one can show that sums , differences , products and quotients of continuous functions are continuous on their domains under appropriate conditions . For example : 
 ( i ) Any polynomial function of two variables and is continuous on . Note that a polynomial of and is a function which is obtained by adding terms of the form where and are non negative integers . 
 ( ii ) A rational function is continuous on as numerator and denominator are continuous on and the denominator satisfies for all . 
 THEOREM 5 Continuity of Composite Functions 
 If is continuous at and is continuous at , then the composite function is continuous at . 
 Investigate the continuity of , . 
 Solution . Since is continuous for all and is continuous for all , is also continuous for all by using Theorem 5 . Similarly , since is continuous for all and is continuous for all , is also continuous for all by using Theorem 5 . ■ 
 The following theorem is an important concept which is used in quite often while dealing with continuous functions . It says that if is a continuous function and has a non zero value at a point , then it continues to be non zero near in the domain of . 
 If is a continuous function at the point and , then has the same sign as near the point . 
 Continuity of Functions of Three Variables 
 The notions of limit and continuity for functions of three or more variables are natural extensions of those just considered . For example , a function is continuous at if 
 The polynomial function is continuous throughout . The rational function 
 is continuous except at the single point . The rational function 
 is continuous except at points on the plane . 
 13 . 2 EXERCISES ( Limits and Continuity of Multivariate Functions ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 2 - Sol . html 
 1 - 7 . Find the limit , if it exists , or show that the limit does not exist . 
 Solution . Let and along the line . 
 For example , for , and for . Therefore the limit does not exists . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 1 . html 
 The limit exists and it is . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - Exs - 3 . html 
 Answer : The limit does not exist . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 3 . html 
 After plotting the function , we know it converges . So the order in and does not effect . 
 sin ( y ^ 3 ) / y ^ 2 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 4 . html 
 Let and along the line . 
 Answer : The limit exists and it is . 
 After plotting the function , we know it converges . Let and along the line . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 6 . html 
 8 . What value of for 
 make the function continuous at . 
 Solution . ⇒ 
 Since , then by Squeeze Theorem . 
 Hence , if we define , then will be continuous at . 
 9 - 10 . Let each of the following functions have the value 0 at the origin . Which of them are continuous at the origin ? Explain your answer . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 7 . html 
 The limit is . Hence , if we define , then is continuous at the origin . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 1 - 9 . html 
 var ( ' x , y , z , w ') 
 11 - 12 . Find and the set on which is continuous . 
 Solution . Since is continuous for all and is continuous for all , is continuous on a set by using Theorem 5 in Section 13 . 2 . 
 http : / / youtu . be / LR 89 Ct 3 cEDY 
 In this section , we will deal with partial derivatives , which is an important concept in multivariate calculus . Partial derivatives are ( essentially ) derivative of one variable functions . Consider defined on a domain , . Let us fix a point . Suppose we fix one of the variables , say , then becomes a function of one variable . Let . If is differentiable at , that is , 
 exists , then we say that has a partial derivative at with respect to . Similarly we can define the partial derivative of with respect to at . See Definition 1 for a generalization of this idea . 
 Let be a function of two variables and . 
 The partial derivative of with respect to is defined as 
 , provided this limit exists . 
 The partial derivative of with respect to may be denoted by ( any of the following ) 
 Similarly , the partial derivative of with respect to may be denoted by 
 Geometric Meaning of the Partial Derivative 
 Let us consider the function . Let be a point on the surface where . We wish to look at the geometric meaning of and . To find , we look at the curve of intersection of the surface and the plane , parallel to the plane . This is the same as saying fix in . Then is the slope of the tangent to the curve of intersection at as the rate change of for at this point . ( See Figure 1 . ) 
 Figure 1 Intersection of the surface with the plane 
 Similarly , is the slope of the tangent of the curve at the point as the rate change of for at this point . ( See Figure 2 . ) 
 Figure 2 Intersection of the surface with the plane 
 Find partial derivatives with respect to and for the following functions at the given point . 
 ( b ) The partial derivatives are 
 Then and . ■ 
 Show that for . 
 Solution . The partial derivatives are 
 Let be a function of three variables and . The partial derivatives of with respect to , with respect to , and with respect to are defined as follows 
 provided the limits on the right hand side exist . 
 In general , if is a function of variables , then the partial derivative of with respect to the variable , , is defined to be 
 To compute , we differentiate with respect to while holding the remaining variables fixed . 
 Find , and if . 
 Solution . , and . 
 print fy 
 Answer : y * e ^( x * y ) * log ( z ) , x * e ^( x * y ) * log ( z ) , e ^( x * y ) / z . ■ 
 Second and Higher Order Partial Derivatives 
 Suppose that has first order partial derivatives and . Note that and are also function of and . Therefore , we can talk about their partial derivatives if they exist . If and have partial derivatives with respect to and , then the following four possibilities exist : 
 These are called second order partial derivatives of . and are called mixed second order partial derivatives . In general , the second order mixed partial derivatives need not be equal . 
 Third order partial derivatives may also be defined , for instance 
 The higher partial derivative may be defined similarly if they exist . 
 be a function of two variables . 
 Verify that and are not define at . 
 Then is continuous on . 
 print fxy 
 print fyx 
 Since two limits are different , is not define at . 
 Actually on except . 
 Then it is easy to show that , , , exist . However because and . ■ 
 This example shows second order partial derivatives and may be different , so the order of differentiation cannot be interchanged arbitrarily . The following theorem provides a sufficient condition for changing the order of differentiation . 
 THEOREM 2 Equality of Mixed Partial Derivatives 
 Suppose and it ' s derivatives , and are defined in a domain containing a point and all are continuous at . Then 
 As a consequence , if and both exist and are continuous , then . If third partial derivatives exist and are continuous , then is true as well . These results holds for all higher order partial derivatives . 
 For , find the second order partial derivatives . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 3 - Exm - 5 . html 
 print fxx 
 print fyy 
 Answer : y ^ 2 * e ^(- x * y ) - sin ( x - y ) 
 x ^ 2 * e ^(- x * y ) - sin ( x - y ) 
 x * y * e ^(- x * y ) - e ^(- x * y ) + sin ( x - y ) 
 x * y * e ^(- x * y ) - e ^(- x * y ) + sin ( x – y ) ■ 
 Show that the partial derivatives and both exist , but the function is not continuous at . 
 Solution . Since the definition of partial derivatives , we have 
 letting along the line , 
 13 . 3 EXERCISES ( Partial Derivatives ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 3 - Sol . html 
 http : / / youtu . be / rSYLp 1 mSMXY 
 1 . Find partial derivatives with respect to and for the function . 
 Answer : - 12 * x * y + 24 * y , - 6 * x ^ 2 + 24 * x 
 2 . Find partial derivatives with respect to and for the function . 
 3 . Find all second order partial derivatives of the function . 
 4 . Find all second order partial derivatives of the function . 
 5 . Take an example of the solution for the . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 2 - 6 . html 
 6 - 8 . Laplace ’ s equation A classical equation of mathematics is Laplace ’ s equation , which arises in both theory and applications . It governs ideal fluid flow , electrostatic potentials , and the steady - state distribution of heat in a conducting medium . In two dimensions , Laplace ’ s equation is 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 3 - Exs - 6 . html 
 7 . for any real number . 
 9 . The volume of a right circular cone of radius and height is . Show that if the height remains constant while the radius changes the volume satisfies . 
 10 . Let . Show that 
 Once again we try to generalize , the de ﬁ nition of differentiability of one variable function at . Recall that is differentiable at , with its derivative , if . Using the definition of limit , this means the following : Given any , for sufficiently small . That is , 
 with the condition that as . We can generalize this definition to a function . In the sequel , we assume that is de ﬁ ned on some open subset of . 
 The analogous requirement with several variables is the definition of differentiability for functions or two ( or more ) variables . 
 The function is differentiable at provided and exist and the change equals 
 where for fixed and , and are functions that depend only on and , with as . A function is differentiable on an open region if it is differentiable at each point of . 
 Several observations are needed here . First , the definition extends to functions of more than two variables . Second , we will show how differentiability is related to linear approximation and the existence of a tangent plane in Section 13 . 7 . Finally , the conditions of the definition are generally difficult to verify . The following theorem may be useful in checking differentiablity . 
 THEOREM 2 Conditions for Differentiability 
 Suppose the function has partial derivatives and defined on an open region and and are continuous on . Then is differentiable at every point of . 
 We close with the analog of Theorem 3 in Section 3 . 1 , which states that differentiability implies continuity . 
 THEOREM 3 Differentiability Implies Continuity 
 If a function is differentiable at , then it is continuous at . 
 Proof . By the definition of differentiability , 
 where as . Because is assumed to be differentiable , as and approach , we see that 
 Also , because , it follows that 
 which implies continuity of at . ■ 
 Discuss the differentiability and continuity of the function 
 Solution . As the definition of a rational function , is continuous and differentiable at all points . The interesting behavior occurs at the origin . It can be shown that if the origin is approached along the line , then 
 Therefore , the value of the limit depends on the direction of approach , which implies that the limit does not exist , and is not continuous at . By Theorem 3 , is not differentiable at . 
 Let ’ s look at the first partial derivatives of at . A short calculation shows that 
 That is , the partial derivatives exist at . 
 Despite the fact that is not differentiable at , its first partial derivatives exist at . Existence of first partial derivatives at a point is not enough to ensure differentiability at that point . As expressed in Theorem 2 , continuity of first partial derivatives is required for differentiability . It can be shown that and are not continuous at . ■ 
 The partial derivative introduced in the previous section tells us how a function changes when we vary one variable and hold the others fixed . Sometimes we wish to know how a function changes when more than one variable is changed . In section 3 . 4 , we dealt with differential of a function at . Recall that 
 In this section we wish to generalize this for the function of two variables . Let be a function of two variables . Assume that has first order partial derivatives and at in some neighborhood of . Let us also assume for convenience that and are also continuous . 
 Let and represent infinitesimal changes in and , respectively , at . Let be the increment of the function . That is , 
 We add and subtract in , and get 
 Using the mean value theorem in the interval , there exists between and such that 
 Similarly using the mean value theorem in the interval , there exists between and such that 
 Using and , in , we get 
 The above equation can be rewritten as follows : there exist , such that 
 Note that as andy , we have and . Thus under the limiting case we have 
 The is called the differential or total differential of . At time in infinitesimal case we write and , then the differential can be written as 
 In case , is a function of three variables then the total differential is given by 
 If , find the total differential of . 
 Solution . By definitions of the total differential , we have 
 The first order partial derivatives of with respect to and are as follows : 
 The base diameter and height of a cylinder are measured as and , respectively , with a possible error in measurement of as much as for each . Estimate maximum relative error and percentage error in the calculated volume of the cylinder . 
 Solution . The volume of a cylinder with base diameter cm and height cm is . Taking natural logarithms on both sides of this equation , we get 
 Now we take the differentials on both sides to have 
 is a maximum when and are positive . Therefore , we take and along with . This gives 
 The maximum relative error in the calculated volume is . 
 Thus the percentage error in the calculated volume is 5 . 25 %. ■ 
 The dimensions of a rectangular cuboid are measured to be 70 cm , 60 cm and 50 cm , with a possible error in measurement as much 0 . 3 cm in each . Calculate differential and increment to estimate the maximum error in the volume of the rectangular cuboid . 
 Solution . Let , and be the dimensions of the rectangular cuboid . 
 Then the volume is . By , since 
 if , , and . 
 Next we will find the increment of as follows 
 Given , we have found the total differential . If we have the total differential , can we find ? 
 Solution . By , . 
 Integrating with respect to , we obtain 
 where is a constant with respect to , that is a function of . Next we differentiate both sides of with respect to , and we have 
 Then . Thus . 
 Integrating with respect to , we have 
 where is a constant . Putting this in , we have 
 13 . 4 EXERCISES ( Differentiability and Total Differential ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 4 - Sol . html 
 http : / / youtu . be / qDmCWBiXbIA 
 1 . Find the total differential of when . 
 2 . If , find the total differential of . 
 3 . The base diameter and height of a right circular cone are measured as 10 cm and 25 cm , respectively , with a possible error in measurement of as much as for each . Estimate maximum relative error and percentage error in the calculated volume of the right circular cone . 
 Solution . Since the volume of right circular cone is , . 
 The maximum relative error is and the percentage error is 3 . 2 %. 
 4 . Find an approximation using total differential . 
 ( 1 ) Find an approximation of when . 
 ( 2 ) Find an approximation of . 
 Solution . ( 1 ) Since and 
 , the linear approximation is 
 ( 2 ) Let . Since 
 the linear approximation is 
 http : / / youtu . be / r 3 dGYL 1 vkEU 
 Let . If and depend on other variables , we would like to find derivatives or partial derivatives of with respect to these independent variables . There are several variations that can occur . For example , and can be function of another variable , say , or and can be functions of two variables and . The Chain Rule for multivariable function helps to find derivative and partial derivatives of compositions of functions . Let us consider the first case , where and are functions of . 
 THEOREM 1 Chain Rule ( One Independent Variable ) 
 If is a function of and , and and are differentiable functions of , then the composition is a differentiable function of and 
 Proof . If and and are both functions of the parameter of , then is the independent variable . Using in Section 13 . 4 , we have 
 When and . By hypothesis and are continuous 
 or , . ( See Figure 1 . ) ■ 
 Let have continuous first order partial derivatives and be a differentiable function of . Then the composition is a differentiable function of and 
 Proof . Using in Section 13 . 4 , we have 
 Taking the limit of the above equation as gives . 
 Taking the limit of the above equation as gives 
 or , . ( See Figure 2 . ) ■ 
 THEOREM 3 Chain Rule ( Two Independent Variables ) 
 Let have continuous first order partial derivatives with respect to and . Let and also have continuous first order partial derivatives with respect to and . Then 
 for some . When as . 
 Now taking the limit of the above equation as gives 
 It may be shown , similarly that 
 Let have continuous first order partial derivatives with respect to , , . Let , and be functions of independent variables and . Also assume that , and have continuous first order partial derivatives with respect to and . Then 
 . ( See Figure 4 . ) 
 and . ( See Figure 5 . ) 
 If where and , find . 
 Using the chain rule . 
 Solution . Let us look at the tree diagram in Figure 6 . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 2 - Exm 6 . html 
 var ( ' u , x , y , z ') 
 log ( x ^ 2 + y ^ 2 + z ^ 2 ) 
 2 * x / ( x ^ 2 + y ^ 2 + z ^ 2 ) 
 2 * y / ( x ^ 2 + y ^ 2 + z ^ 2 ) 
 2 * z / ( x ^ 2 + y ^ 2 + z ^ 2 ) 
 x * cos ( x ) + sin ( x ) 
 - x * sin ( x ) + cos ( x ) 
 Answer : 2 / x ■ 
 If where , find . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 5 - Exm - 3 . html 
 Solution . Using the chain rule , we have 
 If where and , find , , and . 
 Solution . Taking the first order partial derivatives with respect to and of and , we obtain 
 Solving ( 3 ) and ( 4 ) , we have 
 To find , first we can get the following 
 http : / / matrix . skku . ac . kr / cal - lab / m - cal - 13 - 5 - Exm - 4 . html 
 var ( ' r , t , x , y , r _ x , t _ x , r _ y , t _ y ') 
 print r _ x . simplify _ trig ( ) 
 print t _ x . simplify _ trig ( ) 
 cos ( t ) , - sin ( t ) / r 
 print r _ y . simplify _ trig ( ) 
 print t _ y . simplify _ trig ( ) 
 sin ( t ) , cos ( t ) / r 
 z _ y . simplify _ trig ( ) 
 ( 2 * r * sin ( t ) - r * cos ( t ) ) * e ^(- t ) 
 z _ yx . simplify _ trig ( ) 
 -( sin ( t ) * cos ( t ) + 2 * cos ( t ) ^ 2 – 1 ) * e ^(- t ) . ■ 
 http : / / archives . math . utk . edu / visual . calculus / 3 / implicit . 7 / index . html 
 ( Ⅰ ) For the relationship between two variables and , we say that is an implicit function of if we are given an equation 
 Here can be thought of as a single - valued function in a small neighborhood of . If the relationship where exists , then is an explicit function of . The following theorem holds for implicit function , if has continuous first order partial derivatives of with respect to in the neighborhood at the point where . 
 THEOREM 6 Implicit Differentiation 
 Let have continuous first order derivatives on its domain . Suppose that defines as a differentiable function of . Then 
 Proof . We can apply the chain rule to differentiate both sides of the equation with respect to . Since both and are functions of , we get 
 Then , if . ■ 
 Now we suppose that is given implicity as a function 
 by an equation of the form . This means that for all in the domain of . If and are differentiable , then we can apply Chain Rule to differentiate the equation as follows : 
 THEOREM 7 Implicit Differentiation 
 Let be differentiable on its domain and suppose that defines as an differentiable function of and . Then 
 , where . ( See Figure 8 . ) 
 Proof . We assumed that defines implicitly as a function of and . Then . 
 So this equation becomes 
 When and are functions of 
 we obtain the derivative with respect to 
 by solving the system of equations above , we find . Let us assume that the determinant 
 Therefore , from formula , we obtain 
 Answer : ( y * sin ( x ) - sin ( y ) ) / ( x * cos ( y ) + cos ( x ) ) 
 Suppose is an implicit function of and on any rectangular region that satisfy . Find and . 
 Solution . Let . Since 
 from , we obtain 
 print dzdx 
 print dzdy 
 Answer : -( 3 * x ^ 2 * e ^( y + z ) - y * cos ( x - z ) ) / ( x ^ 3 * e ^( y + z ) + y * cos ( x - z ) ) 
 -( x ^ 3 * e ^( y + z ) - sin ( x - z ) ) / ( x ^ 3 * e ^( y + z ) + y * cos ( x - z ) ) 
 If equation , find and . 
 and . ■ 
 13 . 5 EXERCISES ( The Chain Rule ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 5 - Sol . html 
 http : / / youtu . be / vzN 5 By 6 qzvM 
 1 . If where , and , find and at and . 
 2 . Find and if , where . 
 Solution . ( 1 ) , 
 3 . Find of . 
 4 . Find of . 
 Solution . Take a partial derivative of both sides with respect to , then we have 
 5 . Find and when is an implicitly defined function of and in 
 Solution . If 
 6 . Find , of . 
 Solution . because 
 7 . Find when . 
 8 . Let with . 
 ( a ) Show that and exist everywhere . 
 ( b ) Are and continuous at the origin ? Justify your answer . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 2 - 5 . html 
 Solution . ( a ) If , . 
 var ( ' x , y , h ') 
 print ( fx ( x , y ) ) 
 print ( fy ( x , y ) ) 
 -( x ^ 2 * y - y ^ 3 ) / ( x ^ 4 + 2 * x ^ 2 * y ^ 2 + y ^ 4 ) 
 ( x ^ 3 - x * y ^ 2 ) / ( x ^ 4 + 2 * x ^ 2 * y ^ 2 + y ^ 4 ) 
 Thus and exist everywhere . 
 is not continuous at the origin . 
 9 . A function is called a homogeneous function of degree if all the terms in are of degree . In other words , for any parameter . If is a homogeneous function of degree n then show that 
 This is also called Euler ’ s theorem for homogeneous function . 
 10 . Verify the Euler ’ s theorem for the following : 
 11 . If is a homogeneous function of degree in , and then show that 
 12 . Let . Find at , . 
 13 . Let is a homogeneous function of degree in and . If then show that 
 14 . Let . Show that 
 15 . Let . Show that 
 16 . Show that is a solution of for all and assuming that is a constant . 
 17 . If is the solution of the equation 
 with the condition that as , ﬁ nd the values of and . 
 18 . Let , , and . Find . 
 19 . Let , , . Find at . 
 13 . 6 Directional Derivatives and Gradient 
 http : / / youtu . be / o 8 L _ ShRANjo 
 The partial derivatives and are the rates of change of at in the positive - and - directions . Rates of change in other directions are given by directional derivatives . We open this section by de ﬁ ning directional derivatives and then use the Chain Rule to derive a formula for their values in terms of - and - derivatives . Then we study gradient vectors and show how they are used to determine how directional derivatives at a point change as the direction changes , and , in particular , how they can be used to ﬁ nd the maximum and minimum directional derivatives at a point . 
 In this section we introduce a type of derivative , called a directional derivative , that enables us to find the rate of change of a function of two or more variables in any direction . 
 DEFINITION 1 Directional Derivatives 
 Let be a unit vector . The directional derivative of at in the direction of is defined as 
 if the limit exist . 
 1 . Note that if , then and if , then . 
 2 . Consider a curve in the domain of . Note that is the line passing through in the direction of furthermore , and . Then it is easy to see that . 
 Find the directional derivative of at in direction of . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 6 - Exm - 1 . html 
 Answer : - 5 ■ 
 Suppose that is differentiable at and is any unit vector . Then 
 Find the directional derivative if 
 and is the unit vector given by angle . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 6 - Exm - 2 . html 
 Solution . When , the unit vector is 
 Then . ■ 
 print dirdr 
 Answer : - 3 / 2 * sqrt ( 3 ) + 13 / 2 
 For convenience , we define the gradient of a function to be the vector - valued function whose components are the first - order partial derivatives of , as specified in Definition 3 . We denote the gradient of a function by grad or . 
 The gradient of is the vector - valued function 
 provided both first order partial derivatives exist . 
 3 . In view of Theorem 2 , we have 
 Find the gradients of 
 ( i ) http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 6 - Exm - 3 . html 
 Answer : ( x , y ) |--> ( 8 * x - 5 * y , - 5 * x + 6 * y ) 
 Answer : ( x , y ) |--> ( - 2 * x ^ 2 * y / ( x ^ 2 + y ^ 2 + 1 ) ^ 2 + y / ( x ^ 2 + y ^ 2 + 1 ) , - 2 * x * y ^ 2 / ( x ^ 2 + y ^ 2 + 1 ) ^ 2 + x / ( x ^ 2 + y ^ 2 + 1 ) ) 
 Interpretations of the Gradient 
 From Remark 3 , . Using properties of the dot product , we have 
 where is the angle between and . It follows that has its maximum value when , which corresponds to . Therefore , has its maximum value and has its greatest rate of increase when and point in the same direction . Notice that when , the actual rate of increase is . 
 Similarly , when , we have , and has its greatest rate of decrease when and point in opposite directions . The actual rate of decrease is . These observations are summarized as follows : 
 Write these as three observations : 
 1 . The gradient points in the direction of steepest ascent at . 
 2 . The negative of the gradient points is the direction of steepest descent . 
 3 . Notice that when the angle between and is , which means and are orthogonal . 
 ( a ) Find the derivative of at the point in the direction of the unit vector . 
 ( b ) What is the direction in which the function increase most rapidly at in - plane region ? 
 ( c ) Determine the direction of zero change in at ( 3 , 4 ) . 
 Solution . ( a ) Since , 
 hen the directional derivative is 
 ( b ) The function increases most rapidly in the direction of the gradient . The gradient is 
 Its direction is 
 ( c ) The directions of zero change in at ( 3 , 4 ) are the directions orthogonal to . 
 Computing shows that they are perpendicular . 
 Note that Observation 3 says that in the direction orthogonal to the gradient , the function does not change at . Recall the curve , where is a constant , is a level curve , on which function values is constant . Combing these two observations , we conclude that gradient is orthogonal to the line tangent to the level curve through . 
 THEOREM 4 The Gradient and Level Curves 
 Given a function differentiable at , the line tangent to the level curve of at is orthogonal to the gradient . 
 Proof . A level curve of the function is a curve in the - plane of the form , where is a constant . By Theorem 6 in section 13 . 5 , the slope of the line tangent to the level curve is . 
 It follows that any vector that points in the direction of the tangent line at the point is a scalar multiple of the vector 
 At that same point , the gradient points in the direction 
 The dot product of and is 
 which implies that and are orthogonal to each other . ■ 
 An immediate consequence of Theorem 4 is an alternative equation of the tangent line . The curve described by can be viewed as a level curve for a surface . By Theorem 4 , the line tangent to the curve at is orthogonal to . Therefore , if is a point on the tangent line , then , which , when simplified , gives an equation of the line tangent to the curve . 
 Consider the upper sheet of a hyperboloid of two sheets . 
 ( a ) Verify that the gradient at is orthogonal to the corresponding level curve at that point . 
 ( b ) Find an equation of the tangent line to the level curve at . 
 We now compute the gradient : 
 It follows that . The tangent vector and the gradient are orthogonal because 
 ( b ) An equation of the line tangent to the level curve at is 
 Then the tangent line is . ■ 
 Let be a function of three variables de ﬁ ned on some domain in . The notion of directional derivative and gradient of can be de ﬁ ned just by introducing extra component z in definition of two variable function . For the sake of completion , let us write these definitions : 
 Let be a function of three variables de ﬁ ned in a domain . Let and be an unit vector in . Then the directional derivative of at in the direction of is given by 
 if the right hand side limit exists . 
 Again using chain rule it turns out that 
 Let be a function of three variables de ﬁ ned in a domain . Let and possesses continuous first order partial derivative at . Then the gradient of at is de ﬁ ned as 
 Find the directional derivative of the function at in direction of . 
 Solution . We have , , and so that 
 Since and is a unit vector in the indicated direction . From last definition we obtain 
 Suppose has continuous first order partial derivatives . Then the surface 
 is one of the level surfaces of . Let 
 be differentiable curves passing through the point on where , and . For all , we obtain 
 Then using the chain rule , we have 
 where is a tangent vector of the curve . What we conclude from this is that the gradient passing through is perpendicular to the tangent vector of all differentiable curves along . ( See Figure 1 . ) Thus is perpendicular to the tangent vector of all differentiable curves passing through on . Therefore , these lines are perpendicular to and in the plane passing through . 
 Figure 1 is perpendicular to tangent vector of differentiable function at . So tangent vector is on the surface . 
 Find the directional derivative of the function at the point in the direction of the vector . 
 Solution . The unit vector of is 
 The partial derivatives of at are 
 The gradient of at is 
 Hence the directional derivative of at in the direction of is 
 Estimate how much the value of will change if the point moves unit from straight toward . 
 Solution . We first find the directional derivative of at in the direction of the vector . The unit vector of is 
 The change in that results from moving unit away from in the direction of is approximately 
 unit . ■ 
 Plot the contour plot and level surface of 
 http : / / matrix . skku . ac . kr / cal - lab / cal - DirectionalDerivative . html 
 html ( "< i > < b > Directional derivative 시각화 < p ></ p > </ b >" ) 
 var ( ' x , y , t , z ') 
 if show _ surface : 
 if show _ contour : 
 13 . 6 EXERCISES ( Directional Derivatives and Gradient ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 6 - Sol . html 
 http : / / youtu . be / 2 _ 7 TOUuzJoE 
 1 . Find the directional derivative of the function at the point in the direction of the vector . 
 2 . Find the directional derivative of the function at the point in the direction of the vector . 
 , ⇒ , 
 3 . Find the directional derivative of the function at the point in the direction of the vector . 
 4 . If , find . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 2 - 2 . html 
 f . gradient ( ) # Find gradient 
 ( x , y , z ) |--> ( - x / ( x ^ 2 + y ^ 2 + z ^ 2 ) ^( 3 / 2 ) , - y / ( x ^ 2 + y ^ 2 + z ^ 2 ) ^( 3 / 2 ) , - z / ( x ^ 2 + y ^ 2 + z ^ 2 ) ^( 3 / 2 ) ) 
 5 . Find the gradient of at the point . 
 6 . Use the definition of the gradient , assume that and are differentiable function on , and let be a constant . Prove the following gradient rules . 
 Solution . and are differentiable function on 
 7 . Find the gradient of . 
 13 . 7 Tangent Plane 
 http : / / youtu . be / uOf - 5 YHKGI 4 
 In this section , we introduce tangent plane to a surface and look at how to find its equations . Tangent plane is extension of tangent line to a one variable function 
 Recall that if be a differentiable function of one variable , then the equation of the tangent line at is given by 
 A tangent plane is defined to be a plane that is just tangent to a surface ( it touches the surface in only one point ) . 
 To derive the equation of tangent plane , we will use Theorem 4 in section 13 . 6 : tangent plane is orthogonal to the gradient . Let be a differentiable function in its domain . If is surface which is a level surface of and is a point on , then we have proved that the gradient is orthogonal to the tangent line to level curves . ( See Figure 1 . ) 
 Figure 1 Tangent Plane to a surface 
 Thus one can define tangent plane at to as a plane which is perpendicular to , provided . 
 Thus , if and are points on the tangent plane and and are their corresponding position vectors , respectively , a vector equation of tangent plane is 
 The definition of tangent plane is given below : 
 DEFINITION 1 Tangent Plane 
 Let be a point on the level surface where is not . Then an equation of the tangent plane at is 
 Find an equation of the tangent plane to the graph of the sphere at . 
 Solution . By defining , we find that the given sphere is the level surface passing through . Now , 
 It follows the form in Definition 1 that an equation of the tangent plane is 
 See in Figure 2 . 
 Figure 2 Tangent Plane in Example 2 
 The equation of the tangent plane obtained in for a function at a point where can be obtained using Definition 1 the graph of the function can be thought of level surface of , where . In this case we have 
 Hence by Definition 1 , the equation of tangent plane is 
 This implies that the equation of tangent plane is 
 DEFINITION 2 Equation of Tangent Plane 
 Let be a function of two variables having first order continuous partial derivatives in some domain . Let be a point in the domain of . Then equation of tangent plane to the surface at the point is given by 
 Find the tangent plane to the surface at the point . ( See figure 3 . ) 
 Solution . Let . Then and . Since and , the equation of tangent plane is 
 show ( T ( x , y ) ) 
 p + pt + tgt 
 Let be a function having continuous first order partial derivatives . Then the gradient to the surface at is given by 
 We know that the gradient is perpendicular to the tangent plane to the surface at the point . Hence the parametric equations of the normal line is given by 
 In case and , the symmetric equation of the normal line to the surface at can be written as 
 Find the tangent plane and the normal line at the point to the surface 
 Figure 4 A tangent plane and a normal vector to the surface 
 Solution . 1 . The partial derivatives of at are 
 By Definition 1 , the equation of a tangent plane at is 
 By , the equation of normal line is 
 ( ) . ■ 
 Find the tangent plane and the normal line at the point to the surface . 
 Solution . Given equation can be written as 
 Then the equation can be considered as a level surface of 
 Therefore our tangent plane at is the plane perpendicular to the gradient at . Now we have 
 The equation of tangent plane is 
 The equation of normal line is 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 3 - Exm 7 . html 
 print Nline 
 Use sage to ﬁ nd the tangent plane and the normal line to the surface at , . Also plot all these together . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 7 - Exm - 5 . html 
 print " The tangent plane is " 
 show ( tangent _ plane ) 
 print " The normal line is " 
 show ( normal _ line ) 
 Answer : The equation of the Tangent Plane is : 
 The equation of the normal line is : 
 p + pt + tangent + normal 
 13 . 7 EXERCISES ( Tangent Plane ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 7 - Sol . html 
 http : / / youtu . be / GDkE 8 OqUvsk 
 1 . Find the equation of the tangent plane at the point to the surface . 
 ⇒ The equation of the tangent plane is 
 2 . Find the equation of the tangent plane and the normal line at the point to the surface . 
 Hence at the point , the tangent plane is and the normal line is ( that is , ). 
 3 . Find so that all tangent planes to the surface pass through the origin for all . 
 Solution . Let . Then we have that 
 Since every tangent plane intersects the origin , the equation of the tangent plane at a point on the surface should satisfies 
 Simplifying the equation , we obtain 
 4 . Find the tangent plane of the surface at . 
 Due to implicit differentiation , we have 
 In particular , at we have 
 Thus , the tangent plane is . Simplifying it , we have . 
 5 . Find the equation of the tangent plane and the normal line to the surface at . 
 Equation of tangent plane : 
 Equation of normal line : 
 6 . Let be the surface whose equation in cylindrical coordinates is . Find the tangent plane and the normal line to at the point in rectangular coordinates . 
 Solution . The equation of in rectangular coordinates is given by . 
 If we regard as a level surface of 
 at , then the normal vector of the tangent plane is , where and hence 
 Therefore , the equations of the tangent plane is 
 And the normal line is 
 7 . Let if and let 
 In the direction of what unit vectors does the directional derivative of at exist ? 
 Solution . Suppose we wish to compute the directional derivative of in the direction of a unit vector Then 
 and the limit does not exist unless or Hence the directional derivative of at exists in the directions of and or their unit scalar multiples . 
 8 - 9 . Find the tangent plane to the given surface at the indicated point . 
 8 . at . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 3 - 4 . html 
 9 . at . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 3 - 11 . html 
 10 . Find the equations of the tangent plane and normal line at the point to the paraboloid . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 7 - Exs - 10 . html 
 ( 2 * t - 1 , - 2 * t + 1 , - t – 2 ) 
 So tangency normal vector is clear . 
 11 . Find the equations of the tangent plane and the normal line at the point to the surface . 
 Solution . We have , 
 Therefore by the equation of the tangent plane to the surface at the point is 
 By the equation of the normal line at the point to the surface is 
 12 . The surface and meet in an ellipse . Find parametric equations for the line tangent to at the point . 
 Solution . Let be a point on the intersection of these surfaces . Then the tangent line at to the curve is orthogonal to and . In particular , it is parallel to . 
 The components of and the coordinates of give us equations for the line . We have 
 Thus the tangent to the curve at is parallel to the vector . Therefore , its equation is . 
 13 . Sketch a level curve ( ellipse ) of passing through the pont . Find a vector perpendicular to this ellipse at the point . 
 Solution . The value of at the point is . 
 Therefore , the level curve of passing through is , which is an ellipse . Vector perpendicular to this ellipse at the point is . 
 13 . 8 Extrema of Multivariate Functions 
 http : / / youtu . be / oDZUkOEszOQ 
 We have seen how to find extreme values of a single variable function as an application to derivatives . In this section we look at how to find extrema of multivatiate functions as an application to partial derivatives . As with single variable function , concepts developed in this section are useful for practical optimization problems . 
 Local and Global Maximum and Minimum 
 We begin with the definition of local and relative extrema for a function of two variables and . 
 DEFINITION 1 Local Maximum and Minimum 
 ( i ) A point is a local maximum of a function if for all near in the domain of . 
 ( ii ) A point is a local minimum of a function if for all near in the domain of . 
 Let and suppose has a local minimum at . This means there exists positive real numbers such that for all and . Similarly , we can define a local maximum at . 
 When in , is a single variable function . Suppose the first order partial derivatives of exist at the point , then we have since it has the extreme value at . 
 Likewise , when , has the extreme value at and . Therefore , if is a local maximum or local minimum of , then 
 This is in fact , a necessary condition for a point to be a point of local maximum / minimum of . All such points where first order partial derivatives vanish are called critical points of . 
 A critical point of a function is a point in the domain of for which and , or where one of its partial derivatives does not exist . 
 THEOREM 3 Local Extrema 
 If a function has a local extremum at point and the first derivatives exist at this point , then 
 Theorem 3 says that if has a local maximum or local minimum at , then is a critical point of . In general , can be a critical point of even though is not a local maximum or local minimum value . 
 The discussion below / above suggests the next theorems . 
 Let us look at . The graph of is shown in the Figure 1 . 
 The point in Figure 2 , is a critical point which is neither a local maximum nor local minimum . It is easy to see that and implies that and . That means is the only critical point of . However from the graph , it is clear that is not a local maximum / local minimum . In fact if we take any small neighborhood around , there are points and in this neighborhood where and . Because this surface looks like a saddle horse , such a critical point is called a saddle point . 
 Another way of looking at Theorem 3 is the following . Suppose has continuous ﬁ rst order partial derivatives . We know that is the direction in which function increases with maximum speed . Thus , if is a point of local maximum , then at on the surface , , there is no direction in which function can increase . Hence which implies . Similarly , at the point of local minimum . 
 Find all the critical points for . 
 Solution . The first partial derivatives are 
 Hence , and implies 
 Thus the critical point is . ■ 
 Thus , the critical point is . ■ 
 Find critical points of . Plot the surface and its contour lines . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 8 - Exm - 3 . html 
 Here has five critical points and . This is clear from the graph of the surface . Look at Figure 4 . 
 It is also clear from the Figure Give Reference that f has two points of local maximum and two points of local minimum and a saddle point . Look at contour line of the function . Look at Figure 4 . 
 It is clear from the contour lines of ( See Figure 5 . ) that , near the point of local maximum and local minimum , level curves resembles concentric circles . Near saddle points , the level curves are hyperbolic looking shapes . ■ 
 We know that critical points are likely candidates for point of local maximum , local minimum or saddle point . Analogous to second derivative test for function of one variable , there is second derivative test for functions of two variables to classify critical points as local maximum , local minimum or saddle points . 
 The following theorem tells us how to decide the local maximum and local minimum of a given function at a critical point . 
 THEOREM 4 Second Derivatives Test 
 Suppose and has continuous second - order partial derivatives in a neighborhood of . Let 
 ( 1 ) If and , then is a local maximum . 
 ( 2 ) If and , then is a local minimum . 
 ( 3 ) If , then does not have a local maximum or 
 minimum at . The point is a saddle point . 
 In case ( c ) the point is called a saddle point of and the graph of crosses its tangent plane at . 
 If , the test gives no information : could have a local maximum or local minimum at , or could be a saddle point of . 
 To remember the formula for , it ’ s helpful to write it as a determinant : 
 The proof of theorem follows from Taylor ’ s series of two variables function which is given at the end of this section . 
 Classify the critical points of . 
 Solution . Suppose is a critical point of , then 
 Solving the above equation for , , we get three critical points , and . Look at the Figure 6 and Figure 7 . It is clear that has two local maximum and one saddle point . 
 Now let us use the second derivative test to classify these critical points . We have 
 At the critical point , and . That means . Hence is a saddle point . 
 At the critical point , and . That means and . Hence is a local minimum . 
 At the critical point , and . That means and . Hence is also a local minimum . ■ 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 8 - Exm - 4 . html 
 for sol in cpoints : 
 if ( ( sol [ x ] in RR ) and ( sol [ y ] in RR ) ) : 
 show ( [ sol [ x ] , sol [ y ] ] ) 
 Answer : [− 1 , − 1 ] , [ 1 , 1 ] , [ 0 , 0 ] 
 show ( [ fxx ( a , b ) , fyy ( a , b ) , fxy ( a , b ) ] ) 
 print bool ( D > 0 ) 
 print bool ( fxx ( a , b ) > 0 ) 
 print c # local minimum 
 Answer : [ 12 , 12 , − 4 ] , True , True , - 1 
 Since D > 0 and , is a local minimum . 
 print " Since D < 0 , ( 0 , 0 , 1 ) is a saddle point . " 
 Answer : [ 0 , 0 , − 4 ] , False , False , 1 
 Since D < 0 , ( 0 , 0 , 1 ) is a saddle point . 
 Find the local extreme values of . 
 Solution . The domain of is the entire plane and the partial derivatives and exist everywhere . Therefore , local extreme values can occur only when 
 The only possibility is the origin , where the value of is zero . Since is never negative , we see that the origin gives a local minimum . ( See Figure 8 . ) This can be written by using Theorem 4 . We have 
 Since and , is a point of local minimum . ■ 
 Figure 8 The paraboloid has a local minimum value of at the origin . 
 Find the local extreme values of the function 
 Solution . Note that has continuous second order partial derivatives at all points in the plane . The function therefore has extreme values only at the points where and are simultaneously zero . This leads to 
 Therefore , the point is the only point where may have extreme value . We calculate 
 Then has a local maximum at . The value of at this point is . ■ 
 Find the local maximum and minimum of 
 in the domain . 
 From the second equation , or . That is , or , is an integer number . 
 When , from the first equation , , is an integer number . In this case , we have three critical points and in the given domain . 
 And in another case , when , then . So critical points are and in the given domain . We will check that these points are wether a local minimum point or a local maximum point or a saddle point . 
 For , since , is a saddle point . 
 For , since and , is a local minimum . 
 Therefore has a local minimum at and . ■ 
 Find the point on the plane nearest the origin . 
 Solution . Let be any point on the plane . Then the distance from the origin to the plane is . 
 If is minimum value , then is minimum value . So we get 
 Solving the above equations , we get a critical point . 
 Thus has a local minimum at . Hence the nearest point in the plane is . ■ 
 Just see how to implement the user defined function to list all critical points and tabulate along with its type ( nature ) . This appears in the attached sage worksheet . 
 For any continuous function de ﬁ ned on a closed interval of ﬁ nite length , we know should have an absolute maximum and an absolute minimum value on . In this section , we want to determine the analogue of , a function of two variables and . 
 We will study how to find the absolute maximum and the absolute minimum values of a continuous function on a closed and bounded region in the - plane . 
 DEFINITION 5 Absolute Maximum and Minimum 
 ( i ) If for all in the domain of , then has an absolute maximum at . is an absolute maximum value . 
 ( ii ) If for all in the domain of , then has an absolute minimum at . is an absolute minimum value . 
 Analogous to have extreme values at endpoint , a function of two variables can have extreme values on the boundary of the closed set . 
 Absolute Maximum and Minimum Values on a Closed Bounded Set 
 Let be continuous on a closed bounded set in . To find absolute maximum and minimum values of on , 
 1 . Determine the values of at all critical points in . 
 2 . Find the maximum and minimum values of on the boundary of . 
 3 . The greatest function value found in 1 and 2 is the absolute maximum value of on , and the least function value found in 1 and 2 is the absolute minimum value of on . 
 Find the absolute maximum and minimum values of the on the closed set 
 Solution . To find critical points , we have 
 Then is a critical point and so . 
 We now determine the maximum and minimum values of on the boundary of , which is a circle of radius described by the parametric equations 
 Substituting and in terms of into the function , we obtain a new function that gives the values of on the boundary of . Thus 
 The critical points of satisfy 
 or . Therefore , has critical points and , which correspond to the points and . 
 The function values are 
 ( boundary point ) 
 After comparing these values , we can decide that is an absolute minimum value of and is an absolute maximum value of . ■ 
 * Extreme of Implicit function 
 Let be an implicit function and assume that has second order continuous partial derivatives . 
 In order to find extremum of , we will find the point such that 
 Such points are called singular points of . We have 
 If is a point of local maximum or minimum , then since . 
 Let be a extreme point and , then 
 ( i ) If , then has a local minimum at and is called a point of local minimum . 
 ( ii ) If , then has a local maximum at and is called a point of local maximum . 
 Find the extreme values of the implicit function 
 Since , we have . Then two singular points are 
 Since , and . Thus , 
 By Theorem 2 , is the local maximum value of . ■ 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 4 - Exm 1 . html 
 show ( p + pts ) 
 Notice that at the tangent is parallel to axis , but it is not a point of local extremum . Can you see this from the graph ? The is a local maximum . 
 To classify singular points we need to find the sign of . At , . Therefore is not a point of maximum / minimum value . At , . Theorem 6 and the graph shows is a point of local maximum . The maximum of is which occurs at . 
 we get three singular points 
 For and , since , 
 Hence the implicit function has local maximum value at . ■ 
 * Expansion of Taylor ' s Theorem 
 In this section we shall look at Taylor ’ s Theorem . 
 Taylor ’ s Theorem for function of two variables is as follows . 
 Suppose the function has order continuous partial derivatives at . Let and be small enough . 
 Proof . If we let in , 
 which is a function of . This single variable function can be expanded as following by Maclaurin formula : 
 Let in the above formula , 
 Differentiate with respect to , 
 Let in the above equation . 
 Similarly , we can differentiate with respect to , 
 When we replace and with and in Taylor ’ s Theorem for a two - variable function , it is transformed as follows : 
 Placing , in . 
 This is Maclaurin ’ s Theorem for two - variable functions which is a special case of the Taylor Theorem . 
 The case of at in ( Taylor ) Theorem 7 is 
 which is the Mean Value Theorem for the two variable function . 
 Find the Taylor series for the function at the point up to degree . 
 Expand the Taylor series for the function at . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 8 - Exm - 13 . html 
 taylor ( x * y ^ 3 , ( x , 1 ) , ( y , - 1 ) , 4 ) 
 Answer : ( y + 1 ) ^ 3 *( x - 1 ) + ( y + 1 ) ^ 3 - 3 *( y + 1 ) ^ 2 *( x - 1 ) - 3 *( y + 1 ) ^ 2 + 3 *( y + 1 ) *( x - 1 ) - x + 3 * y + 3 ■ 
 Plot the surface . Find Taylor ' s polynomial of degrees 1 , 2 , 3 , and 4 . Plot the surface and graph of these polynomial together and show the Taylors polynomial approximates the function as the degree increases . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 8 - Exm - 14 . html 
 p + p 2 + p 1 + p 3 + p 4 
 13 . 8 EXERCISES ( Extrema of Multivariate Functions ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 8 - Sol . html 
 http : / / youtu . be / FWmk _ MasIjE 
 1 . Let . Find the critical points of and classify them . Solution . Solve and 
 . So we have critical points , or . If , then . If , then we have . 
 The critical points are 
 Next , we consider second order partial derivatives to get , , . 
 Then and thus we obtain at points , . This implies that , are saddle points . 
 At points , , we observe that . 
 Moreover , since , has a local maximum at . On the other hand , due to , has a local minimum at . 
 2 . Find the extreme values of the function when . 
 : critical points 
 , so has a local maximum 
 3 - 4 . Locate the maxima , minima , and saddle points of the functions . 
 http : / / matrix . skku . ac . kr / cal - lab / cal - 12 - 4 - 3 . html 
 http : / / matrix . skku . ac . kr / LA - Lab / ms - 1 . html 
 http : / / matrix . skku . ac . kr / LA - Lab / ms - 2 . html 
 if ( ( sol [ x ] in RR ) and ( sol [ y ] in RR ) ) : 
 print ( ( sol [ x ] , sol [ y ] ) ) 
 ( 0 , 0 ) , ( 1 , 0 ) , (- 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) , (- 1 , 1 ) , ( 0 , - 1 ) , ( 1 , - 1 ) , (- 1 , - 1 ) 
 def extreme ( f , a , b ) : 
 if ( D > 0 ) : 
 if ( f 11 > 0 ) : 
 return " local minimum " 
 if ( f 11 < 0 ) : 
 return " local maximum " 
 return " inconclusive " 
 if ( D < 0 ) : 
 return " saddle point " 
 table . append ( [ ( sol [ x ] , sol [ y ] ) , extreme ( f , a , b ) ] ) 
 Critical Point Type 
 ( 0 , 0 ) saddle point 
 ( 1 , 0 ) local maximum 
 ( - 1 , 0 ) local maximum 
 ( 0 , 1 ) local minimum 
 ( 1 , 1 ) saddle point 
 ( - 1 , 1 ) saddle point 
 ( 0 , - 1 ) local minimum 
 ( 1 , - 1 ) saddle point 
 ( - 1 , - 1 ) saddle point 
 Solution . Try this on your own . 
 5 . Let . Answer the following : 
 ( a ) Find points of local maximum / minimum and a saddle point when . 
 ( b ) Give a condition on for the case when has only one critical point . 
 ( 1 , 1 ) is a saddle point . 
 are points of local minimum . 
 If has only one critical point has a solution and should not have a solution . So . 
 6 . Find maximum value of on . 
 So the critical point is and thus critical value is 
 On , we have and 
 On , we have and , . 
 So the maximum value is 2 . 
 7 . Find the absolute maximum and minimum of in the domain which is a closed triangle made of three points ( 0 . 0 ) , ( 2 , 1 ) , ( 1 , 2 ) . 
 critical point : , 
 ( 2 ) 1 . moves on 
 The absolute maximum , and the absolute minimum on . 
 2 . moves on The absolute maximum , and the absolute minimum on . 
 3 . moves on 
 Hence the absolute maximum is 2 and the absolute minimum is 0 . 
 8 . Find the absolute maximum and minimum values on the disk D : 
 Solution . interior of : 
 If implies 
 Thus , we get the critical points 
 Critical points are 
 onsider , boundary of : 
 Moreover , is smallest when and largest when But 
 Thus on D the absolute maximum of is and the absolute minimum is 
 9 . Find the Taylor series for the function 
 10 . Expand the Maclaurin series for the function . 
 Constrained Optimization ( Lagrange Multiplier ) 
 So far we have dealt with finding maximum or minimum of a function of one or more variables without any constraints . Such problems are called unconstrained optimization problems . However , most of real life problems of maximization or minimization involves certain outside conditions or constraints . Such problems are called constrained optimization problems . This section deals with such problems where the constraints are of equality types . The method of Lagrange multipliers is a powerful tool for solving this class of problems without the need to explicitly solve the conditions and use them to eliminate extra variables . 
 First of all let us consider a minimization / maximization problem of a function of two variables . 
 Find extreme values of subjected to . 
 Let us look at this problem geometrically . We wish to find the maximum and minimum value of on the unit circle . 
 The point at which such a minimum ( maximum ) occurs is called a minimizer ( maximizer ) . 
 Note that , for various values of represent parallels lines . The value of increases as we move from left to right along the line . Thus we are looking for a point on the circle at which the value of is minimum and maximum . 
 Look at the Figure 1 . It appears that this happens when these curves just touch each other , that is , when they have a common tangent line . In particular , the normal vectors ( gradient ) to the level curves of and at the point of maximum or minimum are parallel to each other . 
 This turns out to be the necessary condition for a point to be minimum of maximum in general under certain conditions . 
 Let us consider finding a minimizer of subjected to . 
 Suppose is a point of minimum on the level curve and that the minimum value is . 
 Let be a parameterized curve on with . Since has minimum value at , the composition function has minimum at . Hence 
 In particular , is orthogonal to . We already know that the is orthogonal to the . Thus at the point of minimizer , is parallel to provided . 
 ( We shall see later why do we need . ) 
 More precisely we have the following theorem : 
 Let have continuous first order partial derivatives . Let be a point of minimizer or maximizer of subjected to . Then and are parallel to each other . In particular , if , then there exists a scalar such that . Here is called the Lagrange multiplier . 
 Let be a point at which has maximum or minimum subjected to . Then there exists such that we have 
 The are called the Lagrange conditions . 
 In order to find minimum / maximum of subjected to , we use the following steps : 
 Step 1 . Find all the points that satisfy the Lagrange conditions . 
 Step 2 . Evaluate at all the points obtained in Step 1 . The largest of these values is the maximum value of and the smallest is the minimum value of . 
 Lagrange Conditions in Three Variables 
 If is a point at which has maximum or minimum subjected to with , then there exists such that we have 
 Optimize subjected to . 
 Solution . Look at the Figure 2 in which level curves of and the curve are shown . It is clear from the Figure that there are fours points at which the two curves have common tangents . In particular , we can expect to get four points at which has extreme values . 
 Using the Lagrange necessary conditions , there exits such that we have the following equations 
 From , we have . This mean or . 
 If , then . So the points are . 
 If , then from the second equation , we have . Hence from third equation , we get . Thus the points are . 
 We have four possible points of optimizers , and . Look at the Figure 2 . From this figure , it is clear that at these points , the contours of and have common tangents . Look at the Figure 3 , it is clear that at these points the gradients of and are parallel to each other . 
 The value of at these points are , , . Hence the minimizer occurs and the minimum value is . The maximum occurs at two points 
 with maximum value . ■ 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 9 - Exm - 1 . html 
 var ( ' x , y , lam , t ') 
 print f ( sol [ j ] [ x ] , sol [ j ] [ y ] ) 
 print " The Solution is tabulated as follows " 
 html . table ( [ [ ' x , y , ( x , y ) ']] + [ ( sol [ j ] [ x ] , sol [ j ] [ y ] , 
 Find the extreme values ( maximum and minimum ) of subjected to the constrained . 
 Solution . Look at the Figure 4 which represent the level curves of and the curve . It is clear that there are four points at which the two level curves have common tangents . In particular , we can expect to have four points where has extreme values . 
 Using the Lagrange conditions , if is a solution of this problem then 
 Combining , the first two equations we get . This implies . Therefore , we have four points 
 The value of the function at the first two points is and the value of the function at the last two points is . Thus the maximum value of and occurs at two points , . The minimum value of and it occurs at , . 
 Look at the following Figure 5 which show that the and are parallel at point of extremes . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 9 - Exm - 2 . html 
 var ( ' x , y , lam , t ') 
 html . table ( [ [ "$ x $", "$ y $", "$ f ( x , y ) $"]] + [ ( sol [ j ] [ x ] , sol [ j ] [ y ] , 
 Answer : x y f ( x , y ) 
 − 2 1 − 2 
 2 − 1 − 2 ■ 
 Find the dimensions of a cylindrical tin ( with bottom and top ) made up of a metal sheet to maximize its volume such that the total surface area is . 
 Solution . Let and be the radius of the base and height of the cylinder . Then the total surface area of the metal sheet is and the volume is . Thus mathematically , we can write the above problem as : 
 maximize subjected to . 
 Using the Lagrange conditions , we get 
 After solving we get . Suppose is a solution of the above system of equation , then further using the constrained we obtain the solution 
 , , , . ■ 
 Find the points on the ellipsoid that are closest and farthest from the point . 
 Solution . If is any point in , then its distance from is given by . 
 This means we need to optimize subjected to . Note that is minimizer ( maximizer ) of iff is minimizer ( maximizer ) of . Thus we need to optimize 
 Substituting the values , and from the first three equations in the fourth , we get 
 Solving the above equation ( using sage ) for we get 
 For , the solution ( using sage ) is 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 9 - Exm - 4 . html 
 var ( ' x , y , z , lam ') 
 if ( ( sol [ x ] in RR ) and ( sol [ y ] in RR ) and 
 ( sol [ z ] in RR ) and ( sol [ lam ] in RR ) and 
 ( sol [ lam ] in RR ) ) : 
 We should plot the points along with the surfaces . 
 Now let us consider the following example which explains why do we need , in Theorem 1 . 
 Maximize / Minimize subjected to . 
 var ( ' x , y , lam ') 
 that is a solution of the above problem however , Lagrange method fails to detect this point . As in this case . Look the Figure 6 . 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 9 - Exm - 5 . html 
 Figure 6 ■ 
 Lagrange Multipliers for Two and More Constraints 
 Here we consider problems of minimization or maximization with two constraints . The method can be generalized to more than two constraints in a similar manner . In particular , we assume that we want to find the maximum and minimum values of a function subject to two constraints of the form and . We assume that and are non zero and not parallel ( that is , they are linearly independent ) . 
 Geometrically , this means that we are looking for the extreme values of f that lies on the curve of intersection of the level surfaces and . 
 Suppose , is a point on the curve of intersection of the level surfaces and and at which has an extreme value say . Let us see what this mean in terms of gradients . 
 Since is a point of extremum of , is orthogonal to the tangent vector to the level surface . On the other hand , we also have is orthogonal to the level curve and is orthogonal to the level curve . This means that lies in the plane spanned by two vectors and . That is , there exist scalars and such that 
 We summarize this in the following theorem ( without proof ) . 
 Let , and be three functions having first order continuous partial derivatives . If is a solution of the problem 
 minimize / maximize subjected to and 
 where and are non zero and parallel . Then there exist scalars and such that 
 The scalars and are called Lagrange multipliers . 
 Witting the above conditions ( ) along with the constraints , we get the Lagrange conditions 
 Many time it is continent to define a Lagrangian function 
 Then the Lagrange conditions ( to ) is equivalent to 
 , , and , . 
 The cone is cut by the plane in some curve . Find the point on that is closest to the origin . 
 This problem can be stated as 
 subjected to , . 
 We define the Lagrangian function 
 Using the necessary condition for optimization we get the following 
 Now we need to solve - for , , , . 
 From we get and from we get . 
 Equating the two values of and simplifying we get 
 This implies either or . 
 ( Case I ) when , from , we have 
 and from , we get 
 Using the above tow equations and solving for we get which does not have a real solution . Thus is not possible . 
 ( Case II ) when , from , we have 
 Solving this equation for we get . Now 
 Thus the critical points ( possible optimizers ) are 
 Hence is a point of maximum . 
 Hence is a point of minimum . ■ 
 http : / / matrix . skku . ac . kr / cal - lab / m - Sec 13 - 9 - Exm - 6 . html 
 if ( ( sol [ x ] in RR ) and ( sol [ y ] in RR ) and ( sol [ z ] in RR ) and ( sol [ lam ] in RR ) and ( sol [ mu ] in RR ) ) : 
 [ 1 . 26138721247417 , 1 . 26138721247417 , 1 . 47722557505167 , 5 . 36439079917344 ] 
 [ 6 . 73861278752583 , 6 . 73861278752583 , − 9 . 47722557505166 , 180 . 635609200827 ] 
 13 . 9 EXERCISES ( Lagrange Multiplier ) 
 http : / / matrix . skku . ac . kr / Cal - Book / part 2 / CS - Sec - 13 - 9 - Sol . html 
 http : / / youtu . be / YMGdQWBzyrI 
 1 . In each of the following problems , find the extreme ( maximum and minimum ) values of subjected to the given constrained : 
 ( i ) subject to . 
 ( ii ) subject to . 
 ( iii ) subject to . 
 ( iv ) subject to . 
 ( v ) subjected to 
 ( vi ) subjected to . 
 ( vii ) subjected to . 
 2 . Find the points on the sphere that are closest and farthest from the point . 
 3 . Determine the dimensions of a rectangular box , open at the top having volume pf 32 cubic feets and requiring the least amount of material for its construction . 
 4 . Find the maximum and minimum of 
 on the ellipse given by the intersection of the cylinder and the plane 
 5 . Find the maximum and minimum of 
 on the ellipse given by the intersection of the cylinder and the plane . 
 6 . Find the maximum and ninimum of when . 
 7 . In the following exercises find the extreme values of subjected to the two constraints . 
 8 . ( a ) Use Lagrange multipliers to find the highest and lowest points on the ellipse 
 ( b ) Use Lagrange multipliers to show that the rectangle with maximum area that has a given perimeter is a square . 
 ( c ) Use Lagrange multipliers to show that the triangle with maximum area that has a given perimeter is equilateral . 
 ( d ) Find the maximum and minimum volumes of a rectangular box whose surface area is $ 1200 cm ^ 2 $ and whose total edge length is $. 
 ( e ) Use Sage to ind the maximum of 
 9 . Find the absolute maximum value and minimum value of the function 
 on the ellipse . 
 Substituting in , then 
 The absolute maximum value is 
 The absolute minimum value is 
 10 . Find the absolute maximum value of the function on the ellipse . 
 Substituting in > Substituting in 
 Now the absolute maximum value of the function is 
 Made by Manager : Prof . Sang - Gu Lee and Dr . Jae Hwa Lee and 김응기 . http : / / matrix . skku . ac . kr / sglee / and http : / / matrix . skku . ac . kr / cal - book /