Question 800673
Find possible dimensions for a closed box with volume 256 cubic inches, surface area 352 square inches, and length that is twice the width. 
The width of the box in inches can be x.
Then length = 2x.
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Surface area = 2(lw) + 2(lh) + 2(wh) 
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2[2x*x) + 2xh + x*h] = 352 
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2x^2 + 2xh + xh = 352
2x^2 + 3xh - 352 = 0
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Volume = l*w*h
2x*x*h = 256
2x^2h = 256
x^2h = 128
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Solve for "h" in terms of "x":
h = 128/x^2
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Substitute into the surface area equation.
2x^2 + 3x(128/x^2) - 352 = 0
----
2x^2 + 384/x - 352 = 0
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x = 5.5641
2x = 11.1282
h = 128/x^2 = 4.1345
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Cheers,
Stan H.
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