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 ::<math>C'(x)\,=\,8\frac{1920}{x^2}\,=\,8\left(1\frac{240}{x^2}\right).</math>   ::<math>C'(x)\,=\,8\frac{1920}{x^2}\,=\,8\left(1\frac{240}{x^2}\right).</math> 
     
−  This derivative is zero precisely when <math style="verticalalign: 10%">x=4\sqrt{15}</math>, which occurs when <math style="verticalalign: 20%">y=8\sqrt{15}</math>, and these are the values that will minimize cost. Also, don't forget the units  feet!  +  This derivative is zero precisely when <math style="verticalalign: 8%">x=4\sqrt{15}</math>, which occurs when <math style="verticalalign: 18%">y=8\sqrt{15}</math>, and these are the values that will minimize cost. Also, don't forget the units  feet! 
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Latest revision as of 17:35, 17 May 2015
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 4^{th} 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 $x$ and $y$ as shown in the image, we need to remember the equation for the area of a rectangle:

 $A\,=\,xy.$

However, we need to construct a new function to describe cost:

 $C\,=\,(2+6)x+(2+2)y\,=\,8x+4y.$

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 $A\,=\,xy$, we can solve for $y$ in terms of $x$. Since $480\,=\,xy$, we find that $y=480/x$.

Step 2:

Find an expression for cost in terms of one variable: Now, we can use the substitution from step 1 to find

 $C(x)\,=\,8x+4y\,=\,8x+4\cdot {\frac {480}{x}}\,=\,8x+{\frac {1920}{x}}.$

Step 3:

Find the derivative and its roots: We can apply the power rule termbyterm to find

 $C'(x)\,=\,8{\frac {1920}{x^{2}}}\,=\,8\left(1{\frac {240}{x^{2}}}\right).$

This derivative is zero precisely when $x=4{\sqrt {15}}$, which occurs when $y=8{\sqrt {15}}$, and these are the values that will minimize cost. Also, don't forget the units  feet!

Final Answer:

The cost is minimized when the dimensions are $8{\sqrt {15}}$ feet by $4{\sqrt {15}}$ feet. Note that the side with the most expensive fencing is the shorter one.

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