document.write( "Question 143854This question is from textbook College Algebra
\n" ); document.write( ": PROBLEM: VOLUME OF A BOX:
\n" ); document.write( "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?
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Algebra.Com's Answer #104678 by Earlsdon(6294)\"\" \"About 
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Start by letting the width of the metal sheet be x inches, then the length is x+10 inches.
\n" ); document.write( "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).
\n" ); document.write( "We can write the equation:
\n" ); document.write( "\"V+=+%28x-4%29%28x%2B10-4%29%282%29\" or...
\n" ); document.write( "\"%28x-4%29%28x%2B6%29%282%29+=+832\" Divide both sides by 2 to simplify a bit.
\n" ); document.write( "\"%28x-4%29%28x%2B6%29+=+416\" Perform the multiplication.
\n" ); document.write( "\"x%5E2%2B2x-24+=+416\" Subtract 416 from both sides.
\n" ); document.write( "\"x%5E2%2B2x-440+=+0\" Factor this quadratic equation.
\n" ); document.write( "\"%28x%2B22%29%28x-20%29+=+0\" so that...
\n" ); document.write( "\"x+=+-22\" or \"x+=+20\" Discard the negative solution as the length cannot be a negative quantity.
\n" ); document.write( "\"x+=+20\"
\n" ); document.write( "The original dimensions of the metal sheet are:
\n" ); document.write( "Width = 20 inches and the length = 30 inches.
\n" ); document.write( "Check:
\n" ); document.write( "The volumes is:
\n" ); document.write( "(x-4)(x+6)(2)(2) = 832 Substitute x = 20.
\n" ); document.write( "\"%2820-4%29%2820%2B6%29%282%29+=+832\"
\n" ); document.write( "\"%2816%29%2826%29%282%29+=+832\"
\n" ); document.write( "\"832+=+832\" OK!
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