Question

A farmer wants to build a rectangular pen enclosing an area of 100 square feet. He will use wooden fencing on one side, which costs $20 per foot. He will use a chain-link fence on the 3 other sides, which costs $10 per foot. What should the dimensions of the pen be to minimize the cost?

Answer 1

Answer:

The dimensions of the pen that minimize the cost of fencing are:

$x \approx 12.25 \:ft$

$y \approx 8.17 \:ft$

Step-by-step explanation:

Let $x$ be the width and $y$ the length of the rectangular pen.

We know that the area of this rectangle is going to be $x\cdot y$.The problem tells us that the area is 100 feet, so we get the constraint equation:

$x\cdot y=100$

The quantity we want to optimize is going to be the cost to make our fence. If we have chain-link on three sides of the pen, say one side of length $y$ and both sides of length $x$, the cost for these sides will be

$10(y+2x)$

and the remaining side will be fence and hence have cost

$20y$

Thus we have the objective equation:

$C=10(y+2x)+20y\\C=10y+20x+20y\\C=30y+20x$

We can solve the constraint equation for one of the variables to get:

$x\cdot y=100\\y=\frac{100}{x}$

Thus, we get the cost equation in terms of one variable:

$C=30(\frac{100}{x})+20x\\C=\frac{3000}{x}+20x$

We want to find the dimensions that minimize the cost of the pen, for this reason, we take the derivative of the cost equation and set it equal to zero.

$\frac{d}{dx} C=\frac{d}{dx} (\frac{3000}{x}+20x)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\C'(x)=\frac{d}{dx}\left(\frac{3000}{x}\right)+\frac{d}{dx}\left(20x\right)\\\\C'(x)=-\frac{3000}{x^2}+20$

$C'(x)=-\frac{3000}{x^2}+20=0\\\\-\frac{3000}{x^2}x^2+20x^2=0\cdot \:x^2\\-3000+20x^2=0\\-3000+20x^2+3000=0+3000\\20x^2=3000\\\frac{20x^2}{20}=\frac{3000}{20}\\x^2=150\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{150},\:x=-\sqrt{150}$

Because length must always be zero or positive we take $x=\sqrt{150}$ as only value for the width.

To check that this is indeed a value of $x$ that gives us a minimum, we need to take the second derivative of our cost function.

$\frac{d}{dx} C'(x)=\frac{d}{dx} (-\frac{3000}{x^2}+20)\\\\C''(x)=-\frac{d}{dx}\left(\frac{3000}{x^2}\right)+\frac{d}{dx}\left(20\right)\\\\C''(x)=\frac{6000}{x^3}$

Because $C''(\sqrt{150})=\frac{6000}{\left(\sqrt{150}\right)^3}=\frac{4\sqrt{6}}{3}$ is greater than zero, $x=\sqrt{150}$ is a minimum.

Now, we need values of both x and y, thus as $y=\frac{100}{x}$, we get

$x=\sqrt{150}=5\sqrt{6}=12.25$

$y=\frac{100}{\sqrt{150}}=\frac{10\sqrt{6}}{3}\approx 8.17$

The dimensions of the pen that minimize the cost of fencing are:

$x \approx 12.25 \:ft$

$y \approx 8.17 \:ft$