SOLUTION: Randy is building a rectangular, fenced dog run beside his barn. He has a budget of $1050 for fencing and it costs him $8.75 per meter for the materials. He plans to use the side o
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-> SOLUTION: Randy is building a rectangular, fenced dog run beside his barn. He has a budget of $1050 for fencing and it costs him $8.75 per meter for the materials. He plans to use the side o
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Question 773758: Randy is building a rectangular, fenced dog run beside his barn. He has a budget of $1050 for fencing and it costs him $8.75 per meter for the materials. He plans to use the side of the barn as one side of the fenced area. What are the dimensions of a dog run that maximizes the area Randy can enclose? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Randy is building a rectangular, fenced dog run beside his barn.
He has a budget of $1050 for fencing and it costs him $8.75 per meter for the materials.
He plans to use the side of the barn as one side of the fenced area.
What are the dimensions of a dog run that maximizes the area Randy can enclose?
:
Find out how much fence he can get with $1050,
this is the total distance of the three sides
:
1050/8.75 = 120 meters
therefore
L + 2W = 120
L = (120-2W)
:
Area
A = L*W
Replace L with (120-2W)
A = W(120-2W)
A = -2W^2 + 120W; a quadratic equation
We can find the max area by finding the axis of symmetry x = -b/(2a)
In this equation we have
W =
W = 30 meters width for max area
L = 120-2(30)
L = 60 meters length for max area
therefore the dimension: 60 by 30 meters, that would be 1800 sq/m
