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

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

`x \approx 12.25 \:ft`

`y \approx 8.17 \:ft`

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=(100)/(x)`

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

`C=30((100)/(x))+20x\\C=(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.

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

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

Because length must always be zero or positive we take
`x=√(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.

`(d)/(dx) C'(x)=(d)/(dx) (-(3000)/(x^2)+20)\\\\C''(x)=-(d)/(dx)\left((3000)/(x^2)\right)+(d)/(dx)\left(20\right)\\\\C''(x)=(6000)/(x^3)`

Because
`C''(√(150))=(6000)/(\left(√(150)\right)^3)=(4√(6))/(3)` is greater than zero,
`x=√(150)` is a minimum.

Now, we need values of both x and y, thus as
`y=(100)/(x)`, we get

`x=√(150)=5√(6)=12.25`

`y=(100)/(√(150))=(10√(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`