SOLUTION: PROBLEM: VOLUME OF A BOX: A RECTANGULAR PIECE OF METAL IS 10 IN. LONGER THAN IT IS WIDE. SQUARES WITH SIDES 2 IN LONG ARE CUT FROM THE FOUR CORNERS, AND THE FLAPS ARE FOLDED UPWAR

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Question 143854This question is from textbook College Algebra
: PROBLEM: VOLUME OF A BOX:
A RECTANGULAR PIECE OF METAL IS 10 IN. LONGER THAN IT IS WIDE. SQUARES WITH SIDES 2 IN LONG ARE CUT FROM THE FOUR CORNERS, AND THE FLAPS ARE FOLDED UPWARD TO FORM AN OPEN BOX. IF THE VOLUME OF THE BOX IS 832 IN.^3, WHAT WERE THE ORIGINAL DIMENSIONS OF THE PIECE OF METAL? This question is from textbook College Algebra

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Start by letting the width of the metal sheet be x inches, then the length is x+10 inches.
The volume of the box (832 cu.in.) will be calculated by the length (less 4 inches) times width (less 4 inches) times the height of the box (2 inches).
We can write the equation:
V+=+%28x-4%29%28x%2B10-4%29%282%29 or...
%28x-4%29%28x%2B6%29%282%29+=+832 Divide both sides by 2 to simplify a bit.
%28x-4%29%28x%2B6%29+=+416 Perform the multiplication.
x%5E2%2B2x-24+=+416 Subtract 416 from both sides.
x%5E2%2B2x-440+=+0 Factor this quadratic equation.
%28x%2B22%29%28x-20%29+=+0 so that...
x+=+-22 or x+=+20 Discard the negative solution as the length cannot be a negative quantity.
x+=+20
The original dimensions of the metal sheet are:
Width = 20 inches and the length = 30 inches.
Check:
The volumes is:
(x-4)(x+6)(2)(2) = 832 Substitute x = 20.
%2820-4%29%2820%2B6%29%282%29+=+832
%2816%29%2826%29%282%29+=+832
832+=+832 OK!