# algebra

posted by cantu

I have a piece of cardboard that is twice as long as it is wide .I f I cut a 1-inch by 1-inch 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?

1. Steve

width = w
length = 2w

(w-2)(2w-2)(1) = 40
(w-1)(w-2) = 20

20 = 4x5, so w = 6

The box is 6 x 12

check: cut off the 1" corners and you get a box 4x10

2. Reiny

width --- x
length ---- 2x

after cut-out
width = x-2
length = 2x-2
height = x

volume = x(x-2)(2x-2) = 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 = 3.837-2 = 1.837
length of box =7.674-2 = 5.674
height = 3.837
vol = 1.837x5.674x3.837 =39.99 , not bad

3. Reiny

forget my solution,
I was thinking of an entirely different question.

4. cantu

thank you

5. Steve

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

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