document.write( "Question 23299: John has 300 feet of lumber to frame a rectangular patio. (the perimeter of a rectangle is 2 times length plus two times width). He wants to maximize the area of his patio ( area of a rectangle is length times width). What should the dimensions of the patio be? \n" ); document.write( "
Algebra.Com's Answer #12090 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! Perimeter = 2l+2w = 300 \n" ); document.write( " l+w =150 \n" ); document.write( " w = 150-l \n" ); document.write( "Area =lw \n" ); document.write( " =l(150-l) \n" ); document.write( " =150l-l^2 \n" ); document.write( "This is a quadratic in \"l\" where \n" ); document.write( "a=-1, b=150, c=0 \n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "The maximum is at l=-b/2a = 150/2=75 \n" ); document.write( "Maximum area occurs when l=75 and w=75\r \n" ); document.write( "\n" ); document.write( "Cheers, \n" ); document.write( "stan H. \n" ); document.write( " \n" ); document.write( " |