SOLUTION: a farmer has 500 feet of fencing with which to build a rectangluar live stock pen and wants to enclose the maximum area. okay here is my prob i know how to get the maximum area but

Question 186475: a farmer has 500 feet of fencing with which to build a rectangluar live stock pen and wants to enclose the maximum area. okay here is my prob i know how to get the maximum area but it also says to show a graph of the possibilties and i am stuck i did get help with and understand how to use varibles but i'm just not sure how i get this graphed is ther some fabulous trick i don't know about??i am struggling with this last math project and totally bombed on the last one i just need a few pointers on what to do thank you all so very much
tanya
Answer by rapaljer(4671) (Show Source): You can put this solution on YOUR website!
Let x = width of the pen.

If you have 500 feet of fence, then this is the perimeter, so
2W + 2L = 500

You need to find the length L in terms of x, remembering that x= width.
2x + 2L = 500 You have to solve for L.

Divide both sides by 2 to simplify things a bit:
x + L = 250

Subtract x from each side
L=250-x

So, now you have a formula for Area in terms of x:
A=W*L
A= x*(250-x)

Graph y=x*(250-x)

To find the maximum area, you can graph this equation. I recommend that you graph this from x=0 to about x=250. It can't go past these values for x, right?

Now, (did you say that you know what the maximum is? It's a square, so each side will be 500/4 = 125, and the area will be 125^2 = 15625!!) graph y values from y=0 up to about 16000. The graph in this window should look like this:

This confirms that the maximum occurs at x = 125, and the value there for the area is y = about 15,000 or 16,000. The exact value can be found by substituting x = 125 into the formula A = x(250-x).

R^2
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