Answer to the September 2, 2002 Problem of the Week

Don't Fence Me In!

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A rancher in Colorado has 750 feet of fencing. He wants to enclose a rectangular region and then divide it into four pens with fencing parallel to one side of the rectangle (see diagram below).

What is the largest possible total area of the four pens?

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Solution to Problem:

The answer is 14,062.5 square feet.

You can solve the problem with calculus or with precalculus. I will outline the precalculus solution:

Let w = the total width of the four pens.
Let x = the five lengths needed to enclose the four pens.

Then you can form two equations:
The total fencing is: 5x + 2w = 750
The Area is:         A = xw

Solve the first equation for w:
w = 375 - 2.5x

Now substitute this expression for w in the Area equation:
A = x (375 - 2.5x)

Simplifying,
A = 375x - 2.5 x^2,
which is the equation of a parabola opening downward. The greatest area would be at the vertex of the parabola.

Perhaps, it is easier to see this if you replace A by y.
Then it will be in terms of x and y like you are accustomed to seeing. Just remember y stands for the area:

y = 375x - 2.5 x^2

You can solve this on a graphing calculator. If you do, be sure to change the domain and range!!! (Use something like 50 to 100 for x; and use 12000 to 15000 for y).

You can also solve this by completing the square. This will give you the coordinates of the vertex (x, y), which will give you the length of each pen (x) and the area of the four pens (y), which, of course, is what you are looking for.

y = 375x - 2.5x^2

I would rewrite it as:
2.5 x^2 - 375 x = -y

2.5 (x^2 - 150x ) = -y

2.5 (x^2 - 150x + 5625) = -y + 2.5(5625)

2.5 (x^2 - 150x + 5625) = -y + 14,062.5

2.5 (x - 75)^2 = - (y - 14,062.5)

-2.5 (x - 75)^2 = y - 14,062.5

so the vertex is (75, 14062.5)
or in other words, the largest area is 14,062.5 square feet
when x = 75 feet. By the way, the other dimension
is 187.5 feet (this is the total width of the four pens).

Correctly solved by: