# How do I begin to solve this problem?

• October 9th 2013, 01:26 PM
promack
How do I begin to solve this problem?
The question is as follows:

"A Farmer encloses three adjacent rectangular corrals with 1500 feet of fencing. Determine the length and width that will yield a maximum area."
• October 9th 2013, 01:48 PM
topsquark
Re: How do I begin to solve this problem?
The perimeter of the fence is a rectangle, of length x and width y, so we have 2x + 2y = 1500. The area is A = xy.

How can you put the area of the corral in terms of one variable? How do you go about finding a maximum area?

-Dan
• October 9th 2013, 01:53 PM
promack
Re: How do I begin to solve this problem?
it's 3 adjacent corrals though (imgur: the simple image sharer). Don't I have to somehow account for the fence that goes between the corrals?
• October 9th 2013, 02:01 PM
HallsofIvy
Re: How do I begin to solve this problem?
There are two long fences. Calling each length "x", their lengths total 2x. There are four fences connecting them, two at the ends, two separating the three pens. If we call the length of each "y", their length total 4y. The total fencing is 2x+ 4y= 1500 feet and the area is xy square feet. You want to maximize xy subject to the constraint 2x+ 4y= 1500 feet.

One way to do that is to solve 2x+ 4y= 1500 for x= 750- 2y and write the area as $xy= (750- 2y)y= 1500y- 2y^2$. You can find the maximum by completing the square.
• October 9th 2013, 02:05 PM
SlipEternal
Re: How do I begin to solve this problem?
How are the corrals set up?

Are they adjacent like this:

$\begin{tabular}{|c|c|c|}\hline & & \\ \hline\end{tabular}$

Or like this:

$\begin{tabular}{|c|c|}\hline & \\ \hline \multicolumn{2}{|c|}{ } \\ \hline \end{tabular}$

Should the corrals all be the same size?

Edit: I just saw the picture posted by the OP. I guess this post is moot.
• October 9th 2013, 03:06 PM
HallsofIvy
Re: How do I begin to solve this problem?
Quote:

Originally Posted by HallsofIvy
There are two long fences. Calling each length "x", their lengths total 2x. There are four fences connecting them, two at the ends, two separating the three pens. If we call the length of each "y", their length total 4y. The total fencing is 2x+ 4y= 1500 feet and the area is xy square feet. You want to maximize xy subject to the constraint 2x+ 4y= 1500 feet.

One way to do that is to solve 2x+ 4y= 1500 for x= 750- 2y and write the area as $xy= (750- 2y)y= 1500y- 2y^2$. You can find the maximum by completing the square.

That should be $xy= (750- 2y)y= 750y- 2y^2$, of course.
• October 9th 2013, 04:33 PM
promack
Re: How do I begin to solve this problem?
So I solved it and got the length as 375 feet and the width as 187.5 feet. Am I correct?
• October 9th 2013, 06:24 PM
SlipEternal
Re: How do I begin to solve this problem?
Quote:

Originally Posted by promack
So I solved it and got the length as 375 feet and the width as 187.5 feet. Am I correct?

Yes, looks good to me.