algebra
posted by cantu .
I have a piece of cardboard that is twice as long as it is wide .I f I cut a 1inch by 1inch square from each corner and fold up the resulting flaps ,I get a box with a volume of 40 cubic inches.what are the dimensions of the cardboard?

width = w
length = 2w
(w2)(2w2)(1) = 40
(w1)(w2) = 20
20 = 4x5, so w = 6
The box is 6 x 12
check: cut off the 1" corners and you get a box 4x10 
width  x
length  2x
after cutout
width = x2
length = 2x2
height = x
volume = x(x2)(2x2) = 40
2x^3  6x^2 + 4x = 40
x^3  3x^2 + 2x  20 = 0
tried ±1, ±2, ±4, ±5 and could not find a "nice" solution, so I ran it through
Wolfram ,,,,,,
http://www.wolframalpha.com/input/?i=x%5E3++3x%5E2+%2B+2x++20+%3D+0
and got
x = appr. 3.837
so the cardboard was 3.837 by 7.674
check:
width of box = 3.8372 = 1.837
length of box =7.6742 = 5.674
height = 3.837
vol = 1.837x5.674x3.837 =39.99 , not bad 
forget my solution,
I was thinking of an entirely different question. 
thank you

oddly enough, I did exactly what you did! Then when I saw the messy answer, I reread the problem.
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