Question 1104503

Let the dimensions be L, W, and the area A.

{{{ L = (60-2W)/2 = 30 - W }}}

{{{ A = LW = (30-W)W  = 30W- W^2 }}}
Take the derivative of A with respect to W:
{{{ dA/dW = 30-2W }}}
Set it to zero:
{{{    30-2W = 0  }}} —>  {{{ W=15m }}} —> {{{ L=15m }}}

To show that's a max, note
{{{ d^2A/dW^2 = -2 }}}   which indicates the curve is concave down (hence max at W=L=15)

—
Ans:  The dimensions of the largest rectangle is  15m x 15m
( Notice how it is a square that maximizes the area )
—