Calculus made easy Exercise 9 Question 4 A piece of string 30 inches long has its two ends joined together
and is stretched by 3 pegs so as to form a triangle. What is the largest
triangular area that can be enclosed by the string?

I took P = a+b+c and Area = 1/2 (b * h) And found h in terms of a and b.Substituting this value to the h in the area equation I differentiated the Area wrt to b and equated to zero.Is this the right way to do it? 
Can this be soved only using heron's forumla(not A = 1/2(base*height)?
If you are convinced without proof that there is a triangle of maximal area it is easy to see that the maximal triangle has to be equilateral: Keep two pegs $A$, $B$ fixed and use the third peg as a pencil to draw an arc of an ellipse $\gamma_{AB}$ with foci $A$, $B$. The triangle $ABC$ will have maximal area when $C$ is a minor vertex of $\gamma_{AB}$, and will have strictly smaller area for all other positions of $C$. It follows that the triangle of maximal area having a given perimeter is isosceles over all three sides, whence equilateral.
Look Thompson said that we can solve it by using the hints given on the last line of some page so I'll solve it using that only , using herons formula $A = \sqrt{S(S-a)(S-b)(S-c)}$ , $S$ here is half of perimeter $a,b$ and $c$ are the sides, now u need the maximum value of the part $(S-a)(S-b)(S-c)$. Adding these $(S-a) + (S-b) + (S-c) = 3S - (a+b+c) = 3(15) - (30) = 15$. Now u need to divide 15 in three parts in such a way that u get the maximum value when u multiply the three together, the hint on that page helps us in figuring out that value and its $(n+m+p)/3$ , so $15/3 = 5$. This means that $(S-a) = 5$ , $(S-b) = 5$ and $(S-c)=5$ too. $S=$ half of perimeter = $15$ . sub the values yourself to get $a=b=c=10$ , area of equilateral triangle=$\sqrt{3}/4  \text{(side)}^2$  which gives us the answer