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 #125390 by rajagopalan(174)\"\" \"About 
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Let width of original Al sheet =x
\n" ); document.write( "Length of original al sheet =2x
\n" ); document.write( "Now from the 4 corners cut away 5x5 square pieces
\n" ); document.write( "Then fold the sheet in to an open box
\n" ); document.write( "Length of box=2x-10 ( you have cut 2 pieces 5 cm long)
\n" ); document.write( "Width of Box=x-10
\n" ); document.write( "Height of box=5
\n" ); document.write( "Volume of box=LBH=(2x-10)*(x-10)*(5)
\n" ); document.write( "(2x^2-30x+100)*5=10500
\n" ); document.write( "10x^2-150x+500=10500
\n" ); document.write( "10*(x^2-15x+50)=10500
\n" ); document.write( "x^2-15x+50=1050
\n" ); document.write( "x^2-15x+50-1050=0
\n" ); document.write( "x^2-15x-1000=0
\n" ); document.write( "x^2-40x+25x-1000=0
\n" ); document.write( "x(x-40)+25(x-40)=0
\n" ); document.write( "(x-40)(x+25)=0
\n" ); document.write( "x=-25 or 40
\n" ); document.write( "take positive value 40 as length can not be negative.
\n" ); document.write( "Antwort:
\n" ); document.write( "Width of sheet=40
\n" ); document.write( "Length of sheet=80\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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