SOLUTION: A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 320 feet of fencing and does not fence the side along the street,
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Question 1042718: A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 320 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?
a. 25,600 ft2
b. 19,200 ft2
c. 12,800 ft2
d. 6400 ft2 Found 2 solutions by ikleyn, josmiceli:Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
A developer wants to enclose a rectangular grassy lot that borders a city street for parking.
If the developer has 320 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?
a. 25,600 ft2
b. 19,200 ft2
c. 12,800 ft2
d. 6400 ft2
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You have a rectangle of the length L and the width W and the fence of the length 320 = 2W + L.
And the question is: find the maximum of the area L*W under the given restriction 2W + L = 320.
Then you have L = 320-2W and
area L*W = (320-2W)*W = .
For the general quadratic function f(x) = the min/max is at x = .
In your case the maximum area occurs at W = = = 80 feet. Then L = 320-2W = 320 - 2*80 = 160 feet.
Then the area is equal to
area = L*W = 160*80 = 12800 .
Answer. 128000 ft^2, option c).
You can put this solution on YOUR website! Let = the length of fence parallel to the street
Let = the side perpendicular to the street
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Equation for total length of fencing:
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Let = the area of the lot
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The W-value of the vertex ( in this case it's a peak ), is: , when the general form of the parabola is:
and ft2
The largest area that can be inclosed is 12,800 ft2
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