Question 1043636: A rancher has 1200 feet of fencing to enclose two adjacent rectangular corrals of equal lengths and widths as shown in the figure below. What is the maximum area that can be enclosed in the fencing?
Found 2 solutions by ikleyn, advanced_Learner:
Answer by ikleyn(52754)   (Show Source):
You can put this solution on YOUR website! .
A rancher has 1200 feet of fencing to enclose two adjacent rectangular corrals of equal lengths and widths
as shown in the figure below. What is the maximum area that can be enclosed in the fencing?
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We have no this "figure below", therefore I prepared the Figure
on the right to show how I see and understand the condition.
In the Figure, L means the length and W means the width.
So, we have 3 pieces of fencing of the length L each and 4 pieces
of fencing of the length W each.
Then we have this equation
3L + 4W = 1200,
from which we have W =  .
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Figure.
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Next, the combined area of the two corals is
A = L*2W =  =  =  ,
and we have to find the length L in a way to maximize the area A, i.e. maximize the quadratic function
A =  .
Now let me remind you that, if you have a quadratic function f(x) =  of the general form,
then it reaches the maximum/minimum at x =  .
For our situation, a =  and b = 600.
Therefore, the maximum is at L = -  =  = 200.
Thus the area get a maximum at L = 200 feet.
Then W = =  = 150 feet.
Answer. The area is maximal at L = 200 feet and W = 150 feet.
Then the area of one coral is 200*150 = 30000 square feet.
The combined area of the two corals is twice this value, i.e. 60000 square feet.
The plot below confirms this solution.
Plot f(L) =
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Answer by advanced_Learner(501)  (Show Source):
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