A man wants to construct a rectangular enclosure around his model railroad. Only three sides must be fenced, since his garage wall will form...
the fourth side. If he uses 24 meters of fencing, what is the maximum area possible?
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We want to have the rectangular enclosure with a maximum area and with three sided fencing of 24 meters.
We assume that the two parallel sides of the rectangular eclosure measure x meters each and the remaining third side is 24-2x meters, as the sum of all 3 sides is 24 meters.
Therefore, the area A(x) = product of two adjacent sides of the rectangle.
So A(x) = x*(24-2x) , Or A(x) = 24x-2x^2.
A(x) is maximum for some x = c , if A'(c) = 0 and A" (c) < 0.
Therefore, we differentiate A(x) = 24x-2x^2:
A'(x) = (24x-2x^2)' = 24-4x. A'(x) = 0 gives 24 -4x = 0 which gives x = c = 24/4 = 6.
Therefore x = c = 6.
A"(x) = (24-4x)' = -4. So A"(6) = -4 which is < 0.
Therefore A(x) is maximum for x = c= 6.
So the rectangular enclosure has the maximum area for the rectangle of width = 6 meters and length = 24-2*6 = 12 meters. Therefore, area = 6*12 = 72 sq meter.
We'll write the area of a rectangular shape.
A = width*length
A = y*x
Since only three sides of the rectangular shape must be fenced and he uses 24m of fencing, we'll write the perimeter of the fenced shape:
24 = 2y + x
We'll use symmetric property and we'll get:
x = 24 - 2y (1)
Now, we'll write the area with respect to y:
A(y) = x*y
A(y) = (24 - 2y)*y
A(y) = 24y - 2y^2
When the function A(y) has a maximum, the derivative of the function A(y) is cancelling.
A'(y) = (24y - 2y^2)'
A'(y) = 24 - 4y
A'(y) = 0
24 - 4y = 0
-4y = -24
y = -24/-4
y = 6 meters
We'll substitute y = 6 in (1):
x = 24 - 2y
x = 24 - 12
x = 12 meters
The area is:
A = x*y
A = 12*6
The maximum area is A = 72 square meters.
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