Answer
$x=75\text{ ft}$
$y=187.5\text{ ft}$
Work Step by Step
Let $x$ be the width and $y$ the length.
$P=5x+2y$
a) Imagine a rectangle divided into four pens by adding three internal fences parallel to one side. Try different shapes:
$x=150,y=75\Rightarrow P=825,A=11,250$
$x=100,y=90\Rightarrow P=770,A=9,000$
$x=75,y=100\Rightarrow P=675,A=7,500$
$x=100,y=75\Rightarrow P=725,A=7,500$
It seems the maximum area occurs when fencing is fully used and the rectangle is balanced.
b) We need:
2 vertical fences (left and right): total length = $2y$
5 horizontal fences (top, bottom, and 3 internal dividers): total length = $ 5x$
c) Expression for area:
$A=x\cdot y$
d) Fencing Constraint
Total fencing used:
$5x+2y=750$
e) Area as a Function of One Variable
Solve for $y$:
$2y=750-5x\Rightarrow y=\frac{750-5x}{2}$
Substitute into area:
$A(x)=x\cdot\frac{750-5x}{2}=\frac{750x-5x^2}{2}$
f) Take derivative:
$A'(x)=\frac{750-10x}{2}$
$A'(x)=0\Rightarrow x_1=0,x_2=75$
As $x>0$, the only solution is $x=75$.
Calculate $y$:
$y=\frac{750-5(75)}{2}=187.5$
Maximum area:
$A_{max}=75\cdot 187.5=14,062.5\text{ ft}^2$
