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A company wishes to design a rectangular box that will have a volume of 1125 cubic inches. The bottom and top of the box is a rectangles with a length which is 3 times the width. What should the dimensions be to minimize the surface area of the box?

2007-11-07 06:21:26 · 3 answers · asked by whoo_kidd1 1 in Science & Mathematics Mathematics

3 answers

x = width
3x = length
1125/ 3x^2 = height
Surface area =A = 2*3x^2 +2x(1125/x^2) + 3x(1125/x^2)
A = 6x^2 +2250/x + 3375/x
A = 6x^2 + 5625/x
dA/dx = 12x - 5625/x^2 = 0
12x^3 -5625 =0
x = cuberoot (5625/12) = 7.768 = width
3x = 23.304 = length
1125/3x^2 = 6.215 = height

2007-11-07 06:40:42 · answer #1 · answered by ironduke8159 7 · 0 1

XYZ=1125
Y=3X Putting the value of Y in above Eq 3*X^2*Z =1125
or Z=1125/3*X^2 = 375/X^2
Now SA= 2(X*Y+Y*Z+Z*X) = 2(3*X^2+3*X*Z +Z*X)
= 6X^2 + 8*Z*X
= 6*X^2 + 8*375/X^2*X
=6*X^2 + 3000 / X
Now differentiating w.r.t X
SA'(X) = 12*X -3000/X^2
Put SA'(X) =0
which implies 12*X - 3000/X^2=0
means 12*X = 3000 /X^2
or X^3 = 3000/12 = 250
or X = (250)^(1/3)
Find other values

2007-11-07 14:46:24 · answer #2 · answered by finelearner 2 · 0 0

Let L be the length, W be the width and H be the height.
L=3W
WLH = 3w^2H = 1125 => H = 375/W^2
Area = f(W) = 2WL + 2(W+L)H = 6W^2 + 3000/W
f'(W) = 12W - 3000/W^2 = 0
W = 6.3
f(6.3) = 714.33 in^2
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Ideas to save time: Use the symbols that make more sense when you read them.

2007-11-07 14:41:55 · answer #3 · answered by sahsjing 7 · 0 0

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