# 022 Exam 2 Sample B, Problem 10

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Use calculus to set up and solve the word problem: A fence is to be built to enclose a rectangular region of 480 square feet. The fencing material along three sides cost $2 per foot. The fencing material along the 4th side costs$6 per foot. Find the most economical dimensions of the region (that is, minimize the cost).

Foundations:
As with all geometric word problems, it helps to start with a picture. Using the variables ${\displaystyle x}$ and ${\displaystyle y}$ as shown in the image, we need to remember the equation for the area of a rectangle:
${\displaystyle A\,=\,xy.}$
However, we need to construct a new function to describe cost:
${\displaystyle C\,=\,(2+6)x+(2+2)y\,=\,8x+4y.}$
Since we want to minimize cost, we will have to rewrite it as a function of a single variable, and then find when the first derivative is zero. From this, we will find the dimensions which provide the minimum cost.

Solution:

Step 1:
Express one variable in terms of the other: Since we know that the area is 480 square feet and ${\displaystyle A\,=\,xy}$, we can solve for ${\displaystyle y}$ in terms of ${\displaystyle x}$. Since ${\displaystyle 480\,=\,xy}$, we find that ${\displaystyle y=480/x}$.
Step 2:
Find an expression for cost in terms of one variable: Now, we can use the substitution from step 1 to find
${\displaystyle C(x)\,=\,8x+4y\,=\,8x+4\cdot {\frac {480}{x}}\,=\,8x+{\frac {1920}{x}}.}$
Step 3:
Find the derivative and its roots: We can apply the power rule term-by-term to find
${\displaystyle C'(x)\,=\,8-{\frac {1920}{x^{2}}}\,=\,8\left(1-{\frac {240}{x^{2}}}\right).}$
This derivative is zero precisely when ${\displaystyle x=4{\sqrt {15}}}$, which occurs when ${\displaystyle y=8{\sqrt {15}}}$, and these are the values that will minimize cost. Also, don't forget the units - feet!
The cost is minimized when the dimensions are ${\displaystyle 8{\sqrt {15}}}$ feet by ${\displaystyle 4{\sqrt {15}}}$ feet. Note that the side with the most expensive fencing is the shorter one.