SOLUTION: A rectangular plot of the ground is to be enclosed by a fence and then divided down the middle by another fence. The fence down the middle costs $3 per running foot and the other f
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Question 995443: A rectangular plot of the ground is to be enclosed by a fence and then divided down the middle by another fence. The fence down the middle costs $3 per running foot and the other fence costs 6$ per running foot. If the area of the plot is to be 1800ft² and the cost of the fence is not to exceed $2310, what are the restrictions on the dimensions of the plot?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A rectangular plot of the ground is to be enclosed by a fence and then divided down the middle by another fence.
The fence down the middle costs $3 per running foot and the other fence costs 6$ per running foot.
If the area of the plot is to be 1800ft² and the cost of the fence is not to exceed $2310, what are the restrictions on the dimensions of the plot?
:
Let L = the length of the plot
Let W = the width
Therefore
L*W = 1800
L = , (use this form for substitution)
:
The fence cost
6(2L) + 6(2W) + 3W = 2310
12L + 12W + 3W = 2310
12L + 15W = 2310
Simplify, divide by 3
4L + 5W = 770
Replace L with
4() + 5W = 770
+ 5W = 770
Simplify, divide by 5
+ W = 154
multiply equation by W
1440 + W^2 = 154W
A quadratic equation
W^2 - 154W + 1440 = 0
You can use the quadratic formula; a=1, b=-154, c=1440, but this will factor to:
(W-10)(W-144) = 0
Two solutions
W = 10
W = 144
:
If W = 10 ft, then 1800/10 = 180 ft is the length, for $2310 fence cost
Check the cost: 6(2*10) + 6(2*180) + 3(10) = 2310
:
The other solution W = 144, the Length is 1800/144 = 12.5ft, does not seem reasonable although, it too will cost $2310
6(2*144) + 6(2*12.5) + 3(144) = 2310
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