A triangle's base and height have a sum of 36 cm. Find the base and height of the triangle such that its area is a maximum possible area.

If the height is h and the base is b, then irrespective of the exact geometry the area of the triangle is: S = (1/2)*b*h.
If a sum of the height and the base is fixed, b+h=L, then h = L - b, and S(b) = (1/2)*b*(L-b) = (1/2)*b*L - (1/2)*b^2.
To find the value of b for which S reaches maximum: dS/db = (1/2)*L - b = 0.
b = L/2;
To make sure that b = L/2 is the maximum, double check that the second derivative is negative: d^2S/(db)^2 = -1.

Just as the area of a rectangle is maximized when the rectangle is a square, the area of a triangle is maximized when the base and height are the same.

Here is a non-calculus approach.

Let x=base and y=height

x+y=36\Rightarrow x=36-y

A=\frac{xy}{2}

Make the sub and we get:

A=\frac{x(36-x)}{2}

A=\frac{-x^{2}}{2}+18x

Without calculus, the maximum can be found by using

x=\frac{-b}{2a}

b=18, \;\ a=-1/2

The base:

x=\frac{-18}{2(-\frac{1}{2})}=18

The height is then 36-18=18