Asked by Sam
Suppose that 32 feet of fencing is available to enclose a rectangular garden., one side of which will be against the side of a house. What dimensions of the garden will guarantee a maximum area?
Answers
Answered by
Damon
A = x*y
32 = x+2y
x = 32-2y
A = (32-2y)y = 32y-2y^2
or
y^2-16y = -A/2
you did not say if you are in algebra or calculus so I will do it the algebra way by completing the square
y^2 -16y + (16/2)^2 = -A/2 + 64
(y-8)^2 = -(1/2)(A-128)
y = 8
x = 16
and the area is 128 ft^2
32 = x+2y
x = 32-2y
A = (32-2y)y = 32y-2y^2
or
y^2-16y = -A/2
you did not say if you are in algebra or calculus so I will do it the algebra way by completing the square
y^2 -16y + (16/2)^2 = -A/2 + 64
(y-8)^2 = -(1/2)(A-128)
y = 8
x = 16
and the area is 128 ft^2
Answered by
bobpursley
area=LW= but L=32-2w
area=W(32-2W)=32w-2w^2
but roots of the equation are
0=2w(16-w), w=0, or w=16
since this is a parabola, the max/min will occur at 8 (halfway between the roots).
so w=8, L=16
area=W(32-2W)=32w-2w^2
but roots of the equation are
0=2w(16-w), w=0, or w=16
since this is a parabola, the max/min will occur at 8 (halfway between the roots).
so w=8, L=16
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