Answer
$\frac{40\sqrt {30}}{3}$ $ft$ and $20\sqrt {30}$ $ft$
Work Step by Step
See the figure.
The fencing cost ${$}20$ per linear foot install and the cost of the fencing on the west side will be split with the neighbor, so the farmer's cost $C$ will be
$C$ = $\frac{1}{2}(20x)+20y+20x$ = $20y+30x$
The area $A$ to be enclosed is $8000$ $ft^{2}$
$A$ = $xy$ = $8000$
$y$ = $\frac{8000}{x}$
$C(x)$ = $20\left(\frac{8000}{x}\right)+30x$ = $\frac{160000}{x}+30x$
$C'(x)$ = $-\frac{160000}{x^{2}}+30$
$C'(x)$ = $0$
$30$ = $\frac{160000}{x^{2}}$
$x$ = $\frac{40\sqrt {30}}{3}$
$C''(x)$ = $\frac{320000}{x^{3}}$
$C''(x)$ $\gt$ $0$ for $x$ $\gt$ $0$
We have a minimum for $C$ when $x$ = $\frac{40\sqrt {30}}{3}$ $ft$ and $y$ = $20\sqrt {30}$ $ft$
The minimum cost is
$20(20\sqrt {30})+30\left(\frac{40\sqrt {30}}{3}\right)$ = $800\sqrt {30}$ $\approx$ ${$}4381.78$
