Optimization problem, unsure how to incorporate a couple aspects into formula.

[William021]

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?

___________________________________________________________________________________________

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.

 

[Deleted member 4993]

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?

__View attachment 24926_________________________________________________________________________________________

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.

According to your sketch:

2*x + 6*y = 64…....... and

x * y = A(rea) ........ Continue

 

[HallsofIvy]

William021, you seem to have neglected the condition that the outer fences require two strands of wire while the inner fences require only one. Taking the length, parallel to wall, to be "x" we need 2x wire for the two strands of wire. Taking the width, perpendicular to the wall, to be "y" we need 2 times 2y= 4y for the outer fence and 2 times y= 2y for the inner fence so a total of 2x+ 6y, as Subhotosh Kahn has.

 

[William021]

Thanks for the help guys.