Question 326076: A farmer decides to enclose a rectangular garden, using the side of a barn as one side of the rectangle. What is the mazimum area that the farmer can enclose with 100 ft of fence? What should the dementions of the garden be to give this area?
Answer by nyc_function(2741)   (Show Source):
You can put this solution on YOUR website! By the way, this is an algebra forum and your question is a calculus question.
Let P = perimeter of rectangular garden
P = 100
2L + 2W = 100
2(L + W) = 100
L + W = 100 / 2
L + W = 50
W = 50 - L
Let A = area
Let W = (50 - L)
A = LW
A(L) = L(50 - L)
A(L) = 50L - L^2
We now need to find the derivative of A(L).
A'(L) = 50 - 2L
For stationary points, set A'(L) = 0
50 - 2L = 0
2L = 50
L = 25
When L = 24, A'(L) = 50 - 2(24) = 2
When L = 26, A'(L) = 50 - 2(26) = -2
Since the graph is increasing, then stationary, then decreasing it is a maximum.
A(L) = 50L - L^2
A(25) = 50(25) - 25²
A(25) = 1250 - 625
A(25) = 625 square feet
I hope this helps.
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