Another math question? try without calculus if u can. if not, do it with calculus

This is supposed to be part of precalculus section.

http://sdrv.ms/LY22ZV

The only way I can think of doing this without calc is massive trial and error.

We have the equation for perimeter: 2L + 2W = 45.
And the equation for area: A = L * W

We know that L = (45 - 2W) / 2, so we can use this to alter the equation for area:

A = (45 - 2W) / 2 * W
A = (45/2)W - W^2

In order to maximize the area we take the derivative and set it equal to 0.
dA/dW = 45/2 - 2W = 0
45/2 = 2W
W = 45/4.

Now 2L + 2(45/4) = 45
2L +45/2 = 45.
2L = 45/2.
L = 45/4.

Length = 45/4, Width = 45/4. Area = 126.6ft ^ 2.

Greatest Area will be a square
L = 45/4 and W = 45/4 ft