Question 269703: I thought I new what I was doing, maybe I'm just tired. I created a system of 2 equations to find length and width and tried to substitute and solve. Please go over the steps and equations used.
A rectangular lot whose perimeter is 220 feet is fenced along three sides. An expensive fencing along the lot's length costs $20 per foot, and an inexpensive fencing along the two side widths costs only $8 per foot. The total cost of the fencing along the three sides comes to $2040. What are the lot's dimensions?
-Thank you
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! A rectangular lot whose perimeter is 220 feet is fenced along three sides. An expensive fencing along the lot's length costs $20 per foot, and an inexpensive fencing along the two side widths costs only $8 per foot. The total cost of the fencing along the three sides comes to $2040. What are the lot's dimensions?
Let the length be x
the sum of two lengths will be 2x
The two widths together will be 220-2x
the cost for one length will be $20 * x
and the cost for two widths will be $8* (220-x)
The total cost is $2040
So, 20x+8(220-x)= 2040
20x+1760-8x=2040
12x=2040-1760
12x=280
x= 280/12
x= 23.33 feet which is the length
Substitute the value of x in the term 220-2x
220-2*23.33
220-46.66= 173.34 which is sum of two widths.
dividing by 2 we get 86.67 feet which is measurement of one width.
23.33 feet is the length
86.67 feet which is the width.
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