SOLUTION: A rectangular lot is to be bounded by a fence on three sides and by a wall on the fourth side. Two kinds of fencing will be used with heavy duty fencing selling for $4 a foot on th

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Question 551630: A rectangular lot is to be bounded by a fence on three sides and by a wall on the fourth side. Two kinds of fencing will be used with heavy duty fencing selling for $4 a foot on the side opposite the wall. The two remaining sides will use standard fencing selling for $3 a foot. What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6600?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A rectangular lot is to be bounded by a fence on three sides and by a wall on the fourth side.
Two kinds of fencing will be used with heavy duty fencing selling for $4 a foot on the side opposite the wall.
The two remaining sides will use standard fencing selling for $3 a foot.
What are the dimensions of the rectangular plot of greatest area that can be fenced in at a cost of $6600?
:
The perimeter
P = L + 2W
The cost
4L + 3(2W) = $6600
4L + 6W = $6600
4L = (6600-6W)
divide both sides by 4
L = 1650 - 1.5W
:
The area:
A = L * W
Replace L with (1650-1.5w)
A = W(1650-1.5W)
A quadratic equation
A = -1.5w^2 + 1650w
Find the axis of symmetry: x = -b/(2a); In this equation: a=-1.5; b=1650
w = %28-1650%29%2F%282%2A-1.5%29
w = +550 ft is the width
find the length
L = 1650 - 1.5(550)
L = 825 ft is the length
:
An area 825 by 550 ft gives max area for $6600
:
:
:
See if this checks out
4(825) + 3(2(550)) =
3300 + 3300 = $6600