document.write( "Question 1173321: Dimensions of a piece of sheet metal. A rectangular piece of sheet metal is 4 inches less long than twice the width. A 2-inch square piece is cut at each corner. The sides are then folded up to form an open box whose volume is 256 cubic inches. Find the length and width of the original piece of metal. \n" ); document.write( "

Algebra.Com's Answer #798534 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "Dimensions of a piece of sheet metal. A rectangular piece of sheet metal is 4 inches less long than twice the width.
\n" ); document.write( "A 2-inch square piece is cut at each corner. The sides are then folded up to form an open box whose volume is 256 cubic inches.
\n" ); document.write( "Find the length and width of the original piece of metal.
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\n" ); document.write( "\n" ); document.write( " The setup in the post by @josgarithmetic is INCORRECT.\r
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\n" ); document.write( "\n" ); document.write( " I came to bring the correct solution.\r
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document.write( "Let w  be the width of original sheet\r\n" );
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document.write( "Then the length of sheet is 2w-4  inches.\r\n" );
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document.write( "The formed box with open top,  volume equation\r\n" );
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document.write( "    2*(w-2*2)*(2w-4-2*2) = 256\r\n" );
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document.write( "    (w-4)*(2w-8) = 128\r\n" );
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document.write( "    (w-4)*(w-4) = 64\r\n" );
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document.write( "    (w-4)^2 = 64\r\n" );
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document.write( "     w-4 =  = 8\r\n" );
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document.write( "     w   = 8+4 = 12.\r\n" );
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document.write( "ANSWER.  The original dimensions of the metal sheet are  12 inches (the width) and 2*12-4 = 20 inches (the length).\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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