# Maximum problem???

G

#### Guest

##### Guest
You have a 500 foot roll of fencing, which is to form three sides of a rectangular enclosure (the fourth side is an existing brick wall). What are the dimensions of the enclosure with the maximum area?

I know how to solve it if it was four sides, but I'm not sure how to solve it if it's only for three sides.

#### Unco

##### Senior Member
Hi, angel!

We have
Code:
                         Brick wall
_______________
|              |
Fence, length |              |  Fence, length x
x          |              |
|______________|
Fence,
length y
Three sides just means the length of fence is given by (using the variables I used in the diagram):
x+x+y= 500
2x + y = 500

You're confortable with rearranging this to get one variable the subject and plugging this into the equation for the area (A=xy), and differentiating and setting to zero?

With four sides you had 2y + 2x = 500. There isn't much difference, is there?

#### TchrWill

##### Full Member
If x = the short side and 500 - 2x = the long side, the area becomes
A = x(500 - 2x) = 500x - 2x^2

Taking the first derivitive and setting equal to zero yields dA/dx = 500 - 4x = 0 making x = 125 and the long side 250.

G

#### Guest

##### Guest
Thanks Unco, that really helped me a lot!

TchrWill, I'm not sure what you mean by derivitive, dA/dx or how you got 4x. But then again, I'm a bit slow in math.^^; I found the dimensions by:

2x + y = 500
A = x(500 - 2x)
-2x^2 + 500

x = -500/2(-2)
-500/-4 = 125

y = 500 - 2(125) = 250