SOLUTION: 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_

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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___.
I started by using the first equation given to help me solve the problem+V=%282w%5E2H%29%2F2W%5E2 then got 128%2Fw%5E2=H so... S=2LW+2WH+2LH...2%282w%29w%2B2w%28128%2Fw%5E2%29%2B2%282w%29%28128%2Fw%5E2%29
Then i got 4w%5E2%2B%28256%2Fw%29%2B%28512%2Fw%29i don't know how to simplify this any more. But, i was told i needed to graph it, then whatever the x-intercepts are becomes the answer. Please help?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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)
----
2[2x*x) + 2xh + x*h] = 352
------------
2x^2 + 2xh + xh = 352
2x^2 + 3xh - 352 = 0
==========================
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
=================
Substitute into the surface area equation.
2x^2 + 3x(128/x^2) - 352 = 0
----
2x^2 + 384/x - 352 = 0
---------------
x = 5.5641
2x = 11.1282
h = 128/x^2 = 4.1345
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Cheers,
Stan H.
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