SOLUTION: This is a quadratic word problem.
A farmer wishes to enclose a rectangular region bordering a river using 600 ft. of fencing. He wants to divide the region into two equal parts
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Question 247985: This is a quadratic word problem.
A farmer wishes to enclose a rectangular region bordering a river using 600 ft. of fencing. He wants to divide the region into two equal parts using some of the fence material. What is the maximum area that can be enclosed with the fencing?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
This is a quadratic word problem.
A farmer wishes to enclose a rectangular region bordering a river using 600 ft. of fencing.
He wants to divide the region into two equal parts using some of the fence material.
What is the maximum area that can be enclosed with the fencing?
:
To divide it into two equal parts, we have the fence equation:
2L + 3W = 600
2L = 600 - 3W
divide by 2
L = 300 - 1.5W
;
Area = L*W
Substitute for L
A = W*(300-1.5W)
A = 300W - 1.5W^2
As a quadratic equation
A = -1.5W^2 + 300W
:
max area occurs at the axis of symmetry, Formula for that: x = -b/(2a)
in our equation: W = x; b = 300; a = -1.5
W =
W =
W = +100 ft width for max area
:
Find the area; W=100
A = -1.5(100^2) + 300(100)
A = -1.5(10000) + 30000
A = -15000 + 30000
A = 15000 sq/ft is max area
;
:
Check
Find the length
L = 300 - 1.5(100)
L = 300 - 150
L = 150 ft is the length
:
then: 100 * 150 = 15000 sq/ft is the area
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