
Please Check My Answer
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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=57.143x
A=2(57.143x)(x)
A=114.286x2x^2
A'=114.2864x=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?

Ah! an optimization problem, I love those:
Given:
400=4y+3xProbably 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 200x1.5x^2
Now maximize that function by taking the derivative:
2003x=0
Thus,
x=200/3
Then,
100(3/4)(200/3)=y
Thus,
10050=y
Thus, the solution is
x=200/3 and y=50

Thank you for correcting me. I made my error when solving for y. Tried to simplify too much. Thanks a lot.

I've seen this thread on physicsforums.com. I think many of the users here are users there. :cool:

Yeah I posted this on that forum too because sometimes I don't get an answer if I only post it in one forum.