document.write( "Question 1149561: On a rectangular piece of cardboard with perimeter 10 inches, three parallel and equally spaced creases are made. The cardboard is then folded along the creases to make a rectangular box with open ends. Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of x that maximizes the volume enclosed by this box. Then give the maximum volume. Round your responses to two decimal places. \n" ); document.write( "
Algebra.Com's Answer #770895 by ankor@dixie-net.com(22740)\"\" \"About 
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On a rectangular piece of cardboard with perimeter 10 inches, three parallel and equally spaced creases are made.
\n" ); document.write( " The cardboard is then folded along the creases to make a rectangular box with open ends.
\n" ); document.write( " Letting x represent the distance (in inches) between the creases, use a graphing calculator to find the value of x that maximizes the volume enclosed by this box.
\n" ); document.write( " Then give the maximum volume. Round your responses to two decimal places.
\n" ); document.write( ":
\n" ); document.write( "let L = the length of the cardboard
\n" ); document.write( "let x = dist between creases
\n" ); document.write( "then
\n" ); document.write( "2L + 8x = 10
\n" ); document.write( "simplify divide by 2
\n" ); document.write( "L + 4x = 5
\n" ); document.write( "L = (5-4x)
\n" ); document.write( ":
\n" ); document.write( "Volume
\n" ); document.write( "V = x^2 * L
\n" ); document.write( "Replace L with (5-4x)
\n" ); document.write( "V = x^2(5-4x)
\n" ); document.write( "V = -4x^3 + 5x^2
\n" ); document.write( ":
\n" ); document.write( "Graph this equation
\n" ); document.write( "\"+graph%28+300%2C+200%2C+-1%2C+4%2C+-1%2C+4%2C+-4x%5E3%2B5x%5E2%2C+1.16%29+\"
\n" ); document.write( "Max volume when x = .83 inches
\n" ); document.write( "Find the volume
\n" ); document.write( "V = -4(.83^3) + 5(.83^2)
\n" ); document.write( "V = -2.29 + 3.45
\n" ); document.write( "V = 1.16 cu/in is the max volume (green line)
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