Optimization problem: A farmer has 140 ft of fencing with which....

Well thats what I did and I changed w to y so im at

2x+2y=80
=>2x=80-2y
=> x=40-y
A=(40-y)y=0 ← How did you get A = 0??
40y-2y=0 ← How did you get that from above?
40y=2y
20=y← How did you get that from above?

y is my width. Is this correct?

No...
 
The length of the side of the fencing attached to the barn = x + 60

The width of the fenced area = W

The fenced area = A = (x+60) * W ..........................................................(1)

Perimeter of the rectangle = 2 * [(x+60) + W]

Perimeter of the fence (excluding the part taken up by) = 140 = 2 * [(x+60) + W] - 60

140 = 2 * [(x+60) + W] - 60

2*[x + W] = 80 ........................................................................................(2)

Now continue.....

Since I was wrong, and i changed w to y
Im stuck at 2x+2y=80
 
2) A farmer has 140 ft of fencing with which she will enclose a rectangular region adjacent to a barn 60 ft long. If she insists on using the entire length of the barn as a portion of one side of the rectangle, what is the maximum area that may be enclosed?

This problem is "trickier" than it looks.

I was hoping that you would find it yourself by doing the correct work.

The length of the fencing attached to the barn = x

The length of the side of the fencing not-attached to the barn = x+60

The width of the fenced area = W

The fenced area = A = (x+60) * W ..........................................................(1)

Perimeter of the rectangle = 2 * [(x+60) + W]

Perimeter of the fence (excluding the part taken up by) = 140 = 2 * [(x+60) + W] - 60

140 = 2 * [(x+60) + W] - 60

2*[x + W] = 80 ........................................................................................(2)

x + W = 40

x = 40 - W (assuming that x ≥ 0).............................................................(3)

W = 40 - x (assuming that x ≥ 0).............................................................(3)

Using (3) in (1)

A = (100 - W)*W or A = (x + 60)*(40-x)

to locally maximize area we have

dA/dW = 0 →

100 - 2*W = 0

W = 50...................................................................................................(4)

x = -10 (Using 3)

But that cannot happen. So we plot the function A = (x + 60)*(40-x) and find that the A is maximum at x= 0 (A = 2400 sqft) within the given domain (x ≥ 0)

To investigate the domain x ≤ 0, we understand that the fencing will be used only on three sides.


The length of the side of the fencing not-attached to the barn = L

The width of the fenced area = W

The fenced area = A = L * W ..........................................................(1a)

Perimeter of the rectangle = 2 * [L + W]

Perimeter of the fence (excluding the part taken up by) = 140 = 2 * [W] + L


L = 140 - 2*W(assuming that L ≤ 60)........................................................................................(2a)


Using (2a) in (1a)

A = (140 - 2*W) * W

to locally maximize area we have

dA/dW = 0 →

140 - 4*W = 0

W = 35...................................................................................................(4a)

L = 70 (Using 2a)

But this is beyond the domain of L (L≤ 60)

So we plot the function again and find that the area is maximized within the domain when L = 60 (Area = 2400 sq.ft)
 
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