# Difference between revisions of "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, so we require a total of $5$ pieces of length $y$ . We also need $2$ pieces of length $x$ for each pen, which means there are a total of $4\cdot 2$ pieces required of length $x$ . 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 $20{\sqrt {2}}$ feet by $25/{\sqrt {2}}$ feet.