Area of Parking Lot?

Question

Construct a parking lot on land bordered on one side of the highway. It has 320 feet of fencing that it will use to fence off the other three sides. What should be the dimensions of the lot if the enclosed area is to be maximum?

Answer 1

I start with the assumption that the parking lot is a quadrilateral with one side bordered by the road and as given fence on 3 sides
(Rather the data given in the question suggests it can only be a square or a rectangle as other we shall land up with 3 variables and two equation which cannot be solved)
Let the sides of the rectangle be x and y
the fence is used on 3 sides hence x+x+y = 320 (feet)
or 2x+ y = 320
or y = 320 - 2x --------(a)
Area of the parking lot is
Area A = xy = x (320 - 2x) = 320x - 2.x^2
for the Area to be either minimum or maximum where Area is function of x
(using derivatives)
dA/dx = 320 - 4x = 0 (condition for maximum or minimum value of function is d (function (x)/dx = 0
hence x=80 and and y = 320 - 2 X 80 = 160 feet
before we proceed to calculate the area we find out if this a minimum or maximum area
we find second derivative or derivative of derivative
d(dA/dx)dx is d(320-4x)/dx = - 4 (or minus 4)
The negative value indicates that the area is maximum
(conversely if it was positive the area would have been minimum)
therefore the Area is 80 X 160 = 12800 square ft

Subhash

Answer 2

In this case, you are only fencing off 3 sides of this lot. Let x = the width and y = length. If we assume one of the lengths to be adjacent to the highway, then we are only fencing the two widths and the other length (i.e. 2x+y)

Thus, the constraint is:

320 = 2x+y, solving for y, you get:

y = 320-2x

You know that A=xy, now substitute for y:

A = x(320-2x) = 320x-2x^2. Now differentiate

A' = 320-4x. Set to zero to optimize:

0 = 320-4x -----------> x = 80. Plugging this into the constraint equation, you get:

y = 160.

Therefore, the maximum area of the lot can be a rectangle with dimensions 160 by 80 feet

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Hope this helps

Answer 3

Feliz Natal

Answer 4

Let one side to x, the other y
perimeter
=2[x+y]=320
x+y=160
x=160-y
Area=4xy
A
=4[160-y]*y
=640y-4y^2
A'
=640-8y
=0 for area to be max
y=80
x=80
The parking lot has to be a squre
with side 80 feet

Answer 5

106.67 square feet