SOLUTION: the length of a rectangular piece sheet metal is 3 cm less than twice. a 4cm square is cut from end corner of the metal and the sides are bent up so that the metal forms an open bo

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Question 983189: the length of a rectangular piece sheet metal is 3 cm less than twice. a 4cm square is cut from end corner of the metal and the sides are bent up so that the metal forms an open box. if the volume of the box is 208 cm cube, what are the original dimensions of the metal?
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A mistake is in this work. I have not found it yet.

"length of a rectangular piece sheet metal is 3 cm less than twice. "

Twice what? Width?

L, length
w, WIDTH
c, side of the square cut from each corner
v, volume of result box

These are the facts in the description:
system%28L=-3%2B2w%2Cc=4%2Cv=208%29

You also know that wLc=v.

MAKE SOME EQUATIONS OR EXPRESSIONS
A drawing will help you (but not shown here).

Base area, %28L-2c%29%28w-2c%29
Volume, c%28L-2c%29%28w-2c%29=v




Simplify some first (doing purely in symbols as far as comfortable).
c%28wL-2cw-2cL%2B4c%5E2%29=v
c%28wL-2cL-2cw%2B4c%5E2%29=v
c%28w%282w-3%29-2c%282w-3%29-2cw%2B4c%5E2%29=v, substituted for L
c%282w%5E2-3w-4cw%2B6c-2cw%2B4c%5E2%29=v,
c%282w%5E2-3w-4w-2cw%2B6c%2B4c%5E2%29=v
c%282w%5E2-%283%2B4%2B2c%29w%2B6c%2B4c%5E2%29=v
c%282w%5E2-%287%2B2c%29w%2B6c%2B4c%5E2%29=v
highlight_green%282w%5E2-%287%2B2c%29w%2B6c%2B4c%5E2-v%2Fc=0%29-------You can continue this purely in symbols, or you can substitute NOW for c and for v, and continue the steps from that. You want to then solve for w, either through factoring or through formula for general solution of quadratic formula.

Continueing from the substitution for c and v and simplifying, you should obtain highlight_green%282w%5E2-15w%2B36=0%29.
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You continue this.

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BE CAREFUL. UNKNOWN MISTAKE IN THE ABOVE WORK.

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Reworking separately on paper, still not know where mistake is in the above work, These are two steps found (some steps not included here):
2w%5E2-%283%2B6c%29w%2B6c%2B4c%5E2-v%2Fc=0

Substituting for c and v,
2w%5E2-27w%2B36=0
Factorable!

highlight%28%282w-3%29%28w-12%29=0%29
From which you can reject the first factor, knowing that the value or w from it is impossible. Accept the second factor for w=12. Now, go back and find the corresponding L.