Originally Posted by

**DenisB** The triangle is isosceles (of course).

Assuming b is rectangle's width, then triangle's height = 2b.

Let c = triangle's equal sides.

Then c = sqrt(4b^2 + a^2/4) [1]

So we have a + 2b + 2c = 750

Substitute [1] to get: a + 2b + 2(4b^2 + a^2/4) = 750

Leads to the quadratic 12b^2 + 3000b - 4ab + 1500a - 750^2 = 0

Solving for b:

b = (a - 750 + sqrt(a^2 - 6000a + 2250000)) / 6

I used brute strength to get maximum area 33786.8 where:

a = 231, b = 73.13, c = 186.37

Is the solution given? If so, are my calculation ok?

And perhaps Debsta can confirm...or tell me I'm wrong!!