SOLUTION: This problem is in the review section about the farmer and fencing. A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots

Question 171903This question is from textbook Algebra & Trigonometry
: This problem is in the review section about the farmer and fencing.
A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sided. What is the largest area that can be enclosed?

I really have no clue even where to start by solving this problem, I have searched it and found that we need to maximize the area by using a formula that looks something like(2wx2L=?) but after applying it to my problem I came up with this formula (2Lx3W=10,000. I have no clue if I am even going in the right direction. I used the two lengths because we have two long sides, and 3 widths because there is also the dividing line of fence that has to be accounted for. Any help is appreciated.
Thanks,
KimThis question is from textbook Algebra & Trigonometry

Found 2 solutions by Alan3354, stanbon:
Answer by Alan3354(21583) (Show Source):
You can put this solution on YOUR website!
This problem is in the review section about the farmer and fencing.
A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclosed?
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Farmers always want to complicate things, it seems.
Call W the Width and L the Length with 3 pieces.
The total length will be 2W + 3L
The Area is L*W
so,
A = L*W
2W+3L = 10000
Solve for L
3L = 10000-2W
L = (10000-2W)/3
sub for L into the first eqn
A = L*W
A = W*(10000-2W)/3
A = (10000W - 2W^2)/3
Now, to find the max, set the 1st derivative = 0. I don't know if you know what that means, but...
dA/dW = 10000/3 - 4W/3 = 0
10000 - 4W = 0
W = 2500 meters
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Solve for L
2W+3L = 10000
5000 + 3L = 10000
3L = 5000
L = 1667 meters
Area = 2500*5000/3 =~ 4,166,666.67 sq meters
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Making just a square would have enclosed 5000*5000 = 25,000,000 sq meters, the max area of a rectangle with a given perimeter.

Answer by stanbon(48545) (Show Source):
You can put this solution on YOUR website!
A farmer with 10,000 meters of fencing wants to enclose a rectangular field and then divide it into two plots with a fence parallel to one of the sides. What is the largest area that can be enclosed?
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Draw the picture; there are 3 equal segments of fence (call them "x")
That leaves (10,000-3x) to divide between the two remaining sides;
so each of them is (10,000-3x)/2 meters = (5000-(3/2)x) meters.
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EQUATION:
Area = x(5000-(3/2)x) = 5000x - (3/2)x^2
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maximum occurs when x = -b/2a = -5000/[2(-3/2)] = 5000/3
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Area when x = 5000/3 is as follows:
A(5000/3) = 5000(5000/3) - (3/2) (5000/3)^2
= [(5000/3][5000 - (3/2)(5000/3)]
= [5000/3][2500]
= 4,166,666.67 sq. meters
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Cheers,
Stan H.