### Instruction: A rectangular box has volume 216.  Find the smallest possible surface area of the box.### Response: I need to find the dimensions of the box that minimize its surface area.
Let me label the dimensions as x, y, and z.
Then the volume is x*y*z = 216 and the surface area is 2(x*y + y*z + z*x).
[*, I want to use calculus to find the critical points of the surface area function and then check which one is the minimum.
To do that, I need to eliminate one of the variables using the volume constraint.
I can solve for z in terms of x and y as z = 216/(x*y).
Then I can substitute that into the surface area function and get S(x,y) = 2(x*y + y*216/(x*y) + x*216/(x*y)).
I can simplify this a bit by canceling out some factors and get S(x,y) = 2(x*y + 216/x + 216/y).
Now I can take the partial derivatives of S with respect to x and y and set them equal to zero to find the critical points.
The partial derivative of S with respect to x is 2(y - 216/x^2) and the partial derivative of S with respect to y is 2(x - 216/y^2).
Setting both of these equal to zero, I get y = 216/x^2 and x = 216/y^2.
I can substitute one of these into the other and get x^3 = 216 or x = 6.
Then y = 216/6^2 = 6 and z = 216/(6*6) = 6.
So the critical point is (6,6,6).
To check if this is a minimum or a maximum, I can use the second derivative test.
I need to find the second partial derivatives of S and evaluate them at (6,6,6).
The second partial derivative of S with respect to x is 2(432/x^3) and the second partial derivative of S with respect to y is 2(432/y^3).
Both of these are positive at (6,6,6), which means that S is concave up in both directions and (6,6,6) is a minimum.