document.write( "Question 169927: A rectangular piece of aluminum is to be used to form a box. A 5cm square is to be used from each corner and the ends are to be folded up to form an open box whose volume will be 10500cm ^3. If the original piece of aluminum is twice as long as it is wide, what were the dimensions of the original rectangle? \n" ); document.write( "
Algebra.Com's Answer #125373 by nerdybill(7384)\"\" \"About 
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\n" ); document.write( "Let w = width of aluminum
\n" ); document.write( "then
\n" ); document.write( "2w = length of aluminum
\n" ); document.write( ".
\n" ); document.write( "Box AFTER the square is cut:
\n" ); document.write( "width = w-5-5 = w-10
\n" ); document.write( "length = 2w-5-5 = 2w-10
\n" ); document.write( ".
\n" ); document.write( "Volume of box:
\n" ); document.write( "10500 = (w-10)(2w-10)5
\n" ); document.write( "2100 = (w-10)(2w-10)
\n" ); document.write( "2100 = (w-10)2(w-5)
\n" ); document.write( "1050 = (w-10)(w-5)
\n" ); document.write( "1050 = w^2-5w-10w+50
\n" ); document.write( "1050 = w^2-15w+50
\n" ); document.write( "0 = w^2-15w-1000
\n" ); document.write( ".
\n" ); document.write( "Factoring the right:
\n" ); document.write( "0 = (w+25)(w-40)
\n" ); document.write( ".
\n" ); document.write( "w = {40, -25}
\n" ); document.write( ".
\n" ); document.write( "We can toss out the negative solution leaving us with:
\n" ); document.write( "w = 40 cm (width of rectangle)
\n" ); document.write( ".
\n" ); document.write( "length of rectangle:
\n" ); document.write( "2w = 2(40) = 80 cm (length of rectangle) \n" ); document.write( "
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