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)