document.write( "Question 628074: A rectangular piece of metal is 30 in longer than it is wide. Squares with sides 6 in. long are cut from four corners and the flaps are folded upward to form on open box. If the volume of the box is 594 cubic inches, what were the original dimensions of the piece of metal? \n" ); document.write( "
Algebra.Com's Answer #395351 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "If the sides of the cutout squares are 6\", then the depth of the box when the sides are folded up must be 6 inches. Divide the volume by 6 to get the area of the bottom of the box, 99 square inches.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The original piece of metal had a width of \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since you took 6 X 6 inches out of each corner, subtract 12 inches from each of the width and the length to get expressions for the width and length of the bottom of the box.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Expand the binomials, put the quadratic into standard form, and then solve for \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "John \n" ); document.write( " \n" ); document.write( "My calculator said it, I believe it, that settles it \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |