Question 162680: A homeowner wants to fence a rectangular garden using 120ft of fencing. The side of the garage will be used as one side of the rectangle. Find the dimension for which the area is a maximum.
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Let w = width
and L = length
.
then:
2w + L = 120 (equation 1)
and
area = wL (equation 2)
.
Solve for L in equation 1:
2w + L = 120
L = 120-2w
.
Substitute the above into equation 2:
area = wL
area = w(120-2w)
area = 120w-2w^2
area = -2w^2+120w
area = -2(w^2-60w)
.
The standard parabola equation is:
y= a(x-h)2+k
.
Strategy is manipulate the equation so that it fits that form (to find the vertex) -- do this by completing the square:
area = -2(w^2-60w)
area = -2(w^2-60w+900)+1800
area = -2(w-30)^2+1800
.
(h,k) is the vertex:
(30,1800)
.
Therefore, area is maximized when
w = 30 feet (width)
.
to find the length, we said the length is:
L = 120-2w
L = 120-2(30)
L = 120-60
L = 60 feet (length)
.
Conclusion:
dimension is 30 by 60 feet rectangle
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