Posted by **cantu** on Monday, May 7, 2012 at 8:55am.

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?

- algebra -
**Steve**, Monday, May 7, 2012 at 9:41am
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

- algebra -
**Reiny**, Monday, May 7, 2012 at 9:49am
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

- go with Steve - algebra -
**Reiny**, Monday, May 7, 2012 at 9:50am
forget my solution,

I was thinking of an entirely different question.

- algebra -
**cantu**, Monday, May 7, 2012 at 11:04am
thank you

- algebra - to Reiny -
**Steve**, Monday, May 7, 2012 at 12:48pm
oddly enough, I did exactly what you did! Then when I saw the messy answer, I reread the problem.

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