document.write( "Question 280710: Rick has a rectangular piece of cardboard that is twice as long as it is wide. From each of the 4 corners, Rick cuts a 2in square and turns up the flaps to form an uncovered box. If the volume of the box is 896in^3, what were the dimensions of the original piece of cardboard? \n" ); document.write( "
Algebra.Com's Answer #204020 by checkley77(12844)\"\" \"About 
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X*2X=2X^2 is the original dimensions of the cardboard.
\n" ); document.write( "W*L*H=VOLUME
\n" ); document.write( "W=(x-2*2)=x-4 in.
\n" ); document.write( "L=(2x-2*2)=2x-4 in.
\n" ); document.write( "H=2 in.
\n" ); document.write( "(x-4)(2x-4)*2=896
\n" ); document.write( "2(2x^2-8x-4x+16)=896 divide by 2
\n" ); document.write( "2x^2-12x+16=448
\n" ); document.write( "2x^2-12x+16-448=0
\n" ); document.write( "2x^2-12x-432=0
\n" ); document.write( "2(x^2-6x-216)=0
\n" ); document.write( "2(x-18)(x+12)=0
\n" ); document.write( "x-18=0
\n" ); document.write( "x=18
\n" ); document.write( "18-4=14 one side after corner cuts.
\n" ); document.write( "2*18=36-4=32 the other side after corner cuts.
\n" ); document.write( "Proof:
\n" ); document.write( "14*32*2=896
\n" ); document.write( "896=896\r
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