SOLUTION: A farmer wishes to put a fence around a rectangular field and then divide the field into three rectangular plots by placing two fences parallel to one of the sides. If the farmer c
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Question 945997: A farmer wishes to put a fence around a rectangular field and then divide the field into three rectangular plots by placing two fences parallel to one of the sides. If the farmer can afford only 1000 yards of fencing, what dimensions will give the maximum rectangular area? Found 2 solutions by stanbon, josmiceli:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! A farmer wishes to put a fence around a rectangular field and then divide the field into three rectangular plots by placing two fences parallel to one of the sides. If the farmer can afford only 1000 yards of fencing, what dimensions will give the maximum rectangular area?
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Draw the picture.
There are 4 equal pieces and 2 OTHER equal pieces::
4x + 2y = 1000 yds
2x + y = 500
y = -2x+500
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Area = x*y
A = x(-2x+500)
A = -2x^2 + 500x
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Maximum A occurs when x = -b/(2a) = -500/(2*-2) = 125 yds
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Solve for "y"::
y = -2x+500
y = -2*125+500 = 250 yds
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Cheers,
Stan H.
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You can put this solution on YOUR website! Let = the width of the field
Let = the length of the field
Let = the area of the field
The length of fencing needed is:
(1)
The area is:
(2)
-----------------
(1)
(1)
and
(2)
(2)
The maximum is at
where
and
(1)
(1)
(1)
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The dimensions for maximum area are:
250 x 125
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check:
The length of fencing needed is:
(1)
(1)
(1)
OK
Here's the plot of the area:
at
looks OK
(