# One side of a triangle is 20, and the perimeter is 72. Find the maximum area of the triangle.

The maximum area of the triangle is `120 sqrt(13).` Denote the sides as `a = 20 , ` `b , ` and `c . ` The area can be found by Heron's formula:

`A = sqrt ( p ( p - a ) ( p - b ) ( p - c ) ) `

`p = ( a + b + c ) / 2 = 36 ` is a half of the perimeter (the so-called semi-perimeter). To maximize `A , ` it is sufficient to maximize `A^2 .`

It is given that `a + b + c = 72 . ` In other words, `c = 52 - b . ` In such a way, we can express `A^2 ` as a function of one variable `b ` only:

`A^2 ( b ) = 36 * 52 * ( 36 - b ) ( 36 - ( 52 - b ) ) = 36 * 52 * ( 36 - b ) ( b - 16 ) .`

From this formula we see that `b ` must be between 16 and 36 (which can also be derived from the triangle inequality). This function is a quadratic,

`A^2 ( b ) = 36 * 52 * ( -b^2 + 52 b - 16 * 36 ) ,`

and it is simple to determine its maximum: `b_max = 26 `

`A_max = A ( b_max ) = sqrt ( 36 * 52 * 10 * 10 ) = 120 sqrt ( 13 ) .`

So, the triangle with the maximum area will be isosceles.