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Home Q & A Board Science-Related Homework Help Mathematics
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jackson725 jackson725
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12 years ago
A rectangle has its two lower corners on the x-axis and its two upper corners on the curve y=16-x².  For all such rectangles, what are the dimensions of the rectangle with largest area?
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lia
wrote...
#1
12 years ago
in order for the shape to be a rectangle the lower corners must be at the points (a,0) and (-a,0) for some a>=-4 and a<=4.  Now this gives a length of 2a and a height of 16-a^2 for an area of 2a*(16-a^2) or
32a-2a^3
now this area is maximized when its derivative is zero so we have
32-6a^2=0
6a^2=32
a^2=16/3
a=4/sqrt(3)
so this gives dimensions of
8/sqrt(3) and 32/3 and area of
256/(3sqrt(3))
lia
wrote...
#2
12 years ago
y = 16 - x² is an inverted parabola centers of the y axis.
The rectangles base b corners are at x and -x, so base = 2x.
The height h of the parabola at -a and a is the same: 16 - x²

Area = bh = 2x(16 - x²) = 32x - 2x³

Take the derivative of A and solve for 0
32 - 6x² = 0
6x² = 32
x² = 32/6 = 16/3
x = ± ?(16/3) = ± 4?(1/3) = ± 4?3 / 3

Dimensions are:
b = 2 x 4?3 / 3 = 8?3 / 3
h = 16 - (4?3 / 3)²  = 16 - 16*3/9 = (144 -  48)/9 = 96/9 = 32/3
A = 8?3 / 3 * 32/3 = 256?3 / 9
Answer accepted by topic starter
lialia
wrote...
#3
12 years ago
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