document.write( "Question 1075200: The corners of a piece of paper are cut to make an open top box. The cut corners are 0.5 in by 0.5 in. If the width of the paper is two-thirds the height what, what are the dimensions of the paper if the volume is to be 60 cubic inches? \n" ); document.write( "
Algebra.Com's Answer #689884 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! The corners of a piece of paper are cut to make an open top box. The cut corners are 0.5 in by 0.5 in. If the width of the paper is two-thirds the height what, what are the dimensions of the paper if the volume is to be 60 cubic inches? \n" ); document.write( "------- \n" ); document.write( "width = (2/3)h inches \n" ); document.write( "height = h inches \n" ); document.write( "------ \n" ); document.write( "After cutting the corners:: \n" ); document.write( "width = (2/3)h - 1 inches \n" ); document.write( "height = h - 1 inches \n" ); document.write( "------------------------ \n" ); document.write( "Volume Equation: \n" ); document.write( "h(h-1)((2/3)h-1) = 60 cu in \n" ); document.write( "------------------------------ \n" ); document.write( "h((2/3)h^2 - (5/3)h + 1) = 60 \n" ); document.write( "--- \n" ); document.write( "(2/3)h^3 - (5/3)h^2 + h - 60 = 0 \n" ); document.write( "---------------- \n" ); document.write( "Take the derivative:: \n" ); document.write( "V'(h) = 2h^2 - (10/3)h + 1 \n" ); document.write( "---- \n" ); document.write( "Solve V'(h) = 0 \n" ); document.write( "h = 1.274 inches (height of the paper) \n" ); document.write( "(2/3)h = 0.523 inches (width of the paper) \n" ); document.write( "------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( "-------------- \n" ); document.write( " |