Answer
The dimensions for the plot that would enclose the most area are $\frac{250}{3}~ft$ along the west side and the east side, and $125~ft$ along the south side.
Work Step by Step
Let $x$ be the length of the fence on the west and east. Let $y$ be the length of the fence on the south.
Note that $x \geq 0$ and $y \geq 0$
We can express $y$ in terms of the cost $C$ and $x$:
$C = 10x+20x+20y = 5000$
$y = \frac{5000-30x}{20}$
$y = 250-\frac{3x}{2}$
We can write an expression for the area $A$:
$A = xy = x(250-\frac{3x}{2}) = 250x-\frac{3x^2}{2}$
We can find the point where $A'(x) = 0$:
$A(x) = 250x-\frac{3x^2}{2}$
$A'(x) = 250-3x = 0$
$3x = 250$
$x = \frac{250}{3}$
When $0 \lt x \lt \frac{250}{3}$, then $A'(x) \gt 0$
When $x \gt \frac{250}{3}$, then $A'(x) \lt 0$
Thus, $x= \frac{250}{3}~ft$ is the point where $A$ is a maximum.
We can find the length $y$:
$y =250-\frac{3x}{2}$
$y = 250-\frac{3( \frac{250}{3})}{2}$
$y = 250-\frac{250}{2}$
$y = 125~ft$
The dimensions for the plot that would enclose the most area are $\frac{250}{3}~ft$ along the west side and the east side, and $125~ft$ along the south side.