SOLUTION: A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral. He has 250 feet of fencing available and w
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-> SOLUTION: A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral. He has 250 feet of fencing available and w
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Question 240344: A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral. He has 250 feet of fencing available and wants to use all of it. What is the maximum area of the corral?
There's a picture of it here: http://math.library.wisc.edu/reserves/114/114-nenciu-09spex01.pdf (question #5)
Help? I'm so confused! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A horse breeder wants to construct a corral next to his horse barn that is 50 feet long, using all of the barn as one side of the corral.
He has 250 feet of fencing available and wants to use all of it.
What is the maximum area of the corral?
:
Four sides will be required, 2 widths and 2 lengths: x and (x+50)
The fence equation:
x + (x+50) + 2W = 250
2x + 50 + 2W = 250
2x + 2W = 250 - 50
2x + 2W = 200
simplify divide by 2
x + W = 100
W = 100-x)
:
Area = x * W; replace W with (100-x)
A = x(100-x)
A(x) = -x^2 + 100x; this should answer part (A)
:
The x value for max area, will be the axis of symmetry: x = -b/(2a)
In this equation a=-1; b=100
x =
x =
x = +50; this should answer part(b)
:
Find the dimensions
W = 100 - 50
W = 50' is the width
:
L = x + 50
L = 50 + 50
L = 100'
:
Dimensions for max area: 100 by 50 ft; Part(c)
;
:
Check solutions by finding the total fence length
50 + 100 + 2(50) = 250
