document.write( "Question 62777: Could someone please help me with this. I just don't get algebra, but i'm getting better.
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document.write( "John has 300 feet of lumber to frame a rectangular patio (the perimeter of a rectangle is 2 times length plus 2 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, and show how the maximum area of the patio is calculated from the algebraic equation. \r
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document.write( " Show clearly the algebraic steps which prove your dimensions are the maximum area which can be obtained. Use the vertex formula to find the maximum area.
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Algebra.Com's Answer #43623 by stanbon(75887)![]() ![]() ![]() You can put this solution on YOUR website! 2*length+2*width=300 \n" ); document.write( "Divide thru by 2 to get: \n" ); document.write( "length+width = 150 \n" ); document.write( "Solve for width: \n" ); document.write( "width=150-length \n" ); document.write( "-------- \n" ); document.write( "Let the length be \"x\". \n" ); document.write( "Then the width = \"150-x\" \n" ); document.write( "--------- \n" ); document.write( "Area=(length)(width) \n" ); document.write( "Area=x(150-x) \n" ); document.write( "Area=-x^2+150x \n" ); document.write( "-------- \n" ); document.write( "This is a quadratic with a=-1 and b=150 \n" ); document.write( "Maximum area occurs when x=-b/2a \n" ); document.write( "x=-150/(-2)=75 \n" ); document.write( "-------- \n" ); document.write( "Dimensions are \n" ); document.write( "width = 75 ft. \n" ); document.write( "length = 150-75=75 ft. \n" ); document.write( "The maximum area is 75^2=5625 sq ft. \n" ); document.write( "---------------- \n" ); document.write( "Cheers, \n" ); document.write( "Stan H. \n" ); document.write( " |