document.write( "Question 1146037: A rectangular piece of cardboard is 12 in. longer than it is wide. Squares 3 in. on a side are to be cut from each corner, and then the sides will be folded up to make an open box with a volume of 135 in3. Find the length and width of the piece of cardboard. \n" ); document.write( "
Algebra.Com's Answer #767308 by greenestamps(13203)\"\" \"About 
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\n" ); document.write( "The volume of the box is length times width times height. The height is 3 inches, and the volume is 135 cubic inches. So the length times the width is 135/3 = 45.

\n" ); document.write( "When the 3-inch square pieces are cut out of each corner, the new length is still 12 inches more than the new width.

\n" ); document.write( "So we can solve the problem simply by finding two numbers that differ by 12 and whose product is 45.

\n" ); document.write( "A bit of mental arithmetic shows those two numbers to be 3 and 15. Then, since 3-inch square pieces were cut out of each corner of the piece of cardboard, the dimensions of the piece of cardboard were 3+6=9 and 15+6=21.

\n" ); document.write( "ANSWER: The piece of cardboard was 21x9 inches.

\n" ); document.write( "You can, of course, solve the problem using formal algebra.

\n" ); document.write( "x = width
\n" ); document.write( "x+12 = length

\n" ); document.write( "The volume is 135 after cutting out 3-inch squares from each corner:

\n" ); document.write( "\"3%28x-6%29%28%28x%2B12%29-6%29+=+135\"
\n" ); document.write( "\"3%28x-6%29%28x%2B6%29+=+135\"
\n" ); document.write( "\"%28x-6%29%28x%2B6%29+=+45\"
\n" ); document.write( "\"x%5E2-36+=+45\"
\n" ); document.write( "\"x%5E2+=+81\"
\n" ); document.write( "\"x+=+9\"

\n" ); document.write( "ANSWER:
\n" ); document.write( "width = x = 9
\n" ); document.write( "length = x+12 = 21
\n" ); document.write( "
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