# 022 Sample Final A, Problem 2

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A dairy farmer plans to enclose a rectangular pasture adjacent to a river. To provide enough grass for his cows, the fenced pasture must contain 200 square meters of grass. If no fencing is required along the river, what dimensions will use the smallest amount of 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 enclose a rectangular area, such that we use as little fencing as possible for the three fenced sides. 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:   With the variables declared, we need two fences of length $x$ and one fence of length $y$ . Together, we need a total length of $L=2x+y$ .
On the other hand, we know that the pasture has a fixed area of $200$ square meters. Thus, we also know that $xy=200$ .
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.