009A Sample Final A, Problem 7

7. A farmer wishes to make 4 identical rectangular pens, each with 500 sq. ft. of area. What dimensions for each pen will use the least amount of total fencing?

Foundations:
As a word problem, we must begin by assigning variables in order to construct useful equation(s). As an optimization problem, we will be taking a derivative of one of our equations in order to find an extreme point.

 Solution:

Step 1:
Declare Variables:   We are attempting to find the dimensions of a single pen, such that we use as little fencing as possible for all four pens. Let's use x and y as indicated in the image, and simply call the length of fencing required L.

Step 2:
Form the Equations:   Notice that we need fencing between each of the pens (think "lion-antelope-lion-antelope" if this isn't clear). We require 2 pieces of length x for each pen, and a total of 5 pieces of length y. Together, we need a total length of L = 8x + 5y.
On the other hand, we know that each pen has a fixed area of 500 square feet. Thus, we also know that xy = 500.

Step 3:
Optimize:   Since xy = 500, we also know y = 500/x. Plugging this into our equation for length, we have
$L(x)=8x+5\cdot {\frac {500}{x}}=8x+{\frac {2500}{x}}.$
We now take the derivative to find
$L'(x)=8-{\frac {2500}{x^{2}}}={\frac {8x^{2}-2500}{x^{2}}}.$
The denominator can never be zero, and if we set the numerator to zero we find
$x=\pm {\sqrt {\frac {2500}{8}}}=\pm {\frac {50}{2{\sqrt {2}}}}=\pm {\frac {\,\,25}{\sqrt {2}}}.$
Of course, we can't have negative fencing lengths, so we can ignore the negative root. Finally, we use the area relation to find
$y={\frac {500}{25{\sqrt {2}}}}={\frac {500{\sqrt {2}}}{25}}=20{\sqrt {2}}.$
Thus, the least amount of fencing is used when we size our 500 sq. ft. pens as 20√2 feet by 25/√2 feet.

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