Pat M.'s optimization problem

You've got 750 feet of fence and you want to enclose a rectangular area w/
four corrals like so:

+-----+-----+-----+-----+
+     +     +     +     +
+     +     +     +     +
+     +     +     +     +
+     +     +     +     +
+     +     +     +     +
+     +     +     +     +
+-----+-----+-----+-----+

What's the maximum area you can enclose?

Answer: If you call the length x and the width y we see that 5x+2y=750.

This implies that y=375-(5/2)x

The area is

A=xy=x(375-(5/2)x)=-(5/2)x^2+375x

This graphs out to a "spills water" parabola (https://www.desmos.com/calculator/2bbxki5lka).
The formula that gives the x-coordinate of the point at which a parabola
attains its maximum value is -b/(2a) where a is the coefficient of the x^2
term, b is the coefficient of the linear term. Here a=-(5/2) and b=375.
So -b/2a=-375/(-5)=75.

So A=xy=(75)(375-(5/2)75)=(75)(375/2)=14062.5 ft^2