• Dec 20th 2005, 12:25 PM
Vigo
______y__________y______
l------------l--------------l
l------------l--------------l
l------------l--------------l
l--x---------l--x-----------l x
l------------l--------------l
l------------l--------------l
l___________l____________l

A person has 400 ft. of fencing to maximize the area. What are the dimensions?

400=4y + 3x
A=2xy

57.143=y+x
y=57.143-x

A=2(57.143-x)(x)
A=114.286x-2x^2
A'=114.286-4x=0
114.286=4x
x=28.572

400=4y+3(28.572)
400=4y+85.715
315.286=4y
y=78.571

x=28.572 ft.
y=78.571 ft.

Does this look right?
• Dec 20th 2005, 02:30 PM
ThePerfectHacker
Ah! an optimization problem, I love those:
Given:
400=4y+3x--------Probably x,y>0 are the other conditions
Maximize:
2xy

Since
400=4y+3x
we have by solving for y:
100-.75x=y
Thus, 2xy becomes:
2x(100-.75x) same as 200x-1.5x^2
Now maximize that function by taking the derivative:
200-3x=0
Thus,
x=200/3
Then,
100-(3/4)(200/3)=y
Thus,
100-50=y
Thus, the solution is
x=200/3 and y=50
• Dec 20th 2005, 02:51 PM
Vigo
Thank you for correcting me. I made my error when solving for y. Tried to simplify too much. Thanks a lot.
• Dec 20th 2005, 02:57 PM
Jameson
I've seen this thread on physicsforums.com. I think many of the users here are users there. :cool:
• Dec 20th 2005, 07:35 PM
Vigo
Yeah I posted this on that forum too because sometimes I don't get an answer if I only post it in one forum.