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
Joined
Jul 21, 2005
Messages
1,134
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
Joined
Jul 7, 2005
Messages
856
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
 
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