Answer
The farmer can minimize the cost of the fence by making the length of the field 1500 feet and making the width of the field 1000 feet. Note that the fence that divides the field in half also measures 1000 feet.
Work Step by Step
Let $x$ be the width of the field and let $y$ be the length of the field.
Note that $x \geq 0$ and $y \geq 0$
We can express $y$ in terms of the area $A$ and $x$:
$A = xy$
$y = \frac{A}{x}$
To minimize the cost of the fence, we need to minimize $L$, the length of fence that is used:
$L = 3x+2y$
We can find the point where $L'(x) = 0$:
$L = 3x+2y$
$L = 3x+\frac{2A}{x}$
$L'(x) = 3-\frac{2A}{x^2} = 0$
$\frac{2A}{x^2} = 3$
$x^2 = \frac{2A}{3}$
$x = \sqrt{\frac{2A}{3}}$
$x = \sqrt{\frac{(2)(1.5\times 10^6~ft^2)}{3}}$
$x = 1000~ft$
When $0 \lt x \lt 1000$, then $L'(x) \lt 0$
When $x \gt 1000$, then $L'(x) \gt 0$
Thus, $x=1000~ft$ is the point where $L$ is a minimum.
We can find the length $y$:
$y = \frac{A}{x} = \frac{1.5\times 10^6~ft^2}{1000~ft} = 1500~ft$
The farmer can minimize the cost of the fence by making the length of the field 1500 feet and making the width of the field 1000 feet. Note that the fence that divides the field in half also measures 1000 feet.
