Answer
The dimensions for the plot that would minimize the farmer's cost of the fence are $73.0~ft$ along the west side and the east side, and $109.5~ft$ 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 $A$ and $x$:
$A = xy = 8000$
$y = \frac{8000}{x}$
We can write an expression for the cost $C$:
$C = 10x+20x+20y$
$C = 30x+20(\frac{8000}{x})$
$C = 30x+\frac{160,000}{x}$
We can find the point where $C'(x) = 0$:
$C(x) = 30x+\frac{160,000}{x}$
$C'(x) = 30-\frac{160,000}{x^2} = 0$
$30 = \frac{160,000}{x^2}$
$x^2 = \frac{160,000}{30}$
$x = \sqrt{\frac{16,000}{3}}$
$x = 73.0~ft$
When $0 \lt x \lt 73.0$, then $C'(x) \lt 0$
When $x \gt 73.0$, then $C'(x) \gt 0$
Thus, $x= 73.0~ft$ is the point where $C$ is a minimum.
We can find the length $y$:
$y =\frac{8000}{x}$
$y =\frac{8000}{\sqrt{\frac{16,000}{3}}}$
$y = 109.5~ft$
The dimensions for the plot that would minimize the farmer's cost of the fence are $73.0~ft$ along the west side and the east side, and $109.5~ft$ along the south side.
