# 009A Sample Final A, Problem 7

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 ${\displaystyle x}$ and ${\displaystyle y}$ as indicated in the image, and simply call the length of fencing required ${\displaystyle L}$.
Step 2:
Form the Equations:   Notice that we need fencing between each of the pens, so we require a total of ${\displaystyle 5}$ pieces of length ${\displaystyle y}$. We also need ${\displaystyle 2}$ pieces of length ${\displaystyle x}$ for each pen, which means there are a total of ${\displaystyle 4\cdot 2}$   pieces required of length ${\displaystyle x}$. Together, we need a total length of ${\displaystyle L=8x+5y}$.
On the other hand, we know that each pen has a fixed area of ${\displaystyle 500}$ square feet. Thus, we also know that ${\displaystyle xy=500}$.
Step 3:
Optimize:   Since ${\displaystyle xy=500}$, we also know ${\displaystyle y=500/x}$. Plugging this into our equation for length, we have
${\displaystyle L(x)=8x+5\cdot {\frac {500}{x}}=8x+{\frac {2500}{x}}.}$
We now take the derivative to find
${\displaystyle L'(x)=8-{\frac {2500}{x^{2}}}={\frac {8x^{2}-2500}{x^{2}}}.}$
The denominator can never be zero, and if we set the numerator to zero we find
${\displaystyle x=\pm {\sqrt {\frac {2500}{8}}}=\pm {\frac {50}{2{\sqrt {2}}}}=\pm {\frac {\,\,25}{\sqrt {2}}}.}$
Of course, we can't have negative fencing lengths, so we can ignore the negative root. Finally, we use the area relation to find
${\displaystyle y={\frac {500}{25{\sqrt {2}}}}={\frac {500{\sqrt {2}}}{25}}=20{\sqrt {2}}.}$
Thus, the least amount of fencing is used when we size our ${\displaystyle 500}$ sq. ft. pens as ${\displaystyle 20{\sqrt {2}}}$ feet by ${\displaystyle 25/{\sqrt {2}}}$ feet.