William021
New member
- Joined
- Feb 6, 2021
- Messages
- 4
You have 64 feet of wire to enclose three pens. One side is a wall that needs no fence.The outside fencing (thick lines) requires 2 strands of wire. The inside dividers (thin lines) require 1 strand of wire. What values for x and y will create a fence that encloses the maximum total area for the pens?
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What I did so far is this:
The perimeter is compose of x+4y (there are 4 vertical lengths because there are three pens), and one width of wire because the other side needs no fence.
x+4y=64
constraint: x=64-4y
objective function : since a=lw, or xy for this purpose, and because x=64-4y, I get this:
a=(64-4y)y
a=64y-4y^2
I plug this in and get the global maximum at 8, 256.
Its not correct, and I have a feelings its because I'm not considering the middle part of the problem which stipulates how many strands of wire I need.
___________________________________________________________________________________________
What I did so far is this:
The perimeter is compose of x+4y (there are 4 vertical lengths because there are three pens), and one width of wire because the other side needs no fence.
x+4y=64
constraint: x=64-4y
objective function : since a=lw, or xy for this purpose, and because x=64-4y, I get this:
a=(64-4y)y
a=64y-4y^2
I plug this in and get the global maximum at 8, 256.
Its not correct, and I have a feelings its because I'm not considering the middle part of the problem which stipulates how many strands of wire I need.