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Rectangular problem .

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by Md.Nazrul Islam » Wed Apr 04, 2012 7:45 pm
The interior of a rectangular cartoon is designed by a certain manufacturer to have a volume of x cubic feet and a ratio of length to width to height is 3:2:2.In terms of 'x' what is the height of the cartoon in feet .
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by Anurag@Gurome » Wed Apr 04, 2012 8:01 pm
Md.Nazrul Islam wrote:The interior of a rectangular cartoon is designed by a certain manufacturer to have a volume of x cubic feet and a ratio of length to width to height is 3:2:2.In terms of 'x' what is the height of the cartoon in feet .
Let us assume that the length = 3n, width = 2n and height = 2n
Then volume = length * width * height = 3n * 2n * 2n = 12(n^3) = x cubic ft
Solving 12(n^3) = x, we get
n^3 = x/12
n = (x/12)^(1/3)

Therefore, height of the carton = 2 * n = [spoiler]2 * (x/12)^(1/3)[/spoiler]
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by gmatmath » Fri Apr 06, 2012 1:39 am
Given the ratio of length to width to height = 3:2:2.
so, Let the length'l' = 3k, width'w' = 2k and height'h' = 2k
Volume = lxwxh
x = 3k*2k*2k
x = 12k^3
==> k = (x/12)^(1/3)
so, height 'h' = 2k = 2[(x/12)^(1/3)
or h = two times the cuberoot of (x/12)
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