document.write( "Question 636308: The sum of the lengths of the legs of a right triangle is 40cm. What is the largest possible area for such triangle? \n" ); document.write( "

Algebra.Com's Answer #400893 by Edwin McCravy(20055) $\"\"$ $\"About$

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The sum of the lengths of the legs of a right triangle is 40cm. What is the largest possible area for such triangle?
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document.write( "Area = ·base·height\r\n" );
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document.write( "Let the area by y, one leg be x and the other leg be 40-x, then\r\n" );
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document.write( "   y = x(40-x)\r\n" );
document.write( "   y = 40x - x²\r\n" );
document.write( "   y = 20x - x²\r\n" );
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document.write( "Write that in the form \r\n" );
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document.write( "   y = ax² + bx + c\r\n" );
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document.write( "   y = x² + 20x + 0\r\n" );
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document.write( "Since the coefficient of x² is negative, we know that the \r\n" );
document.write( "parabola opens downward and therefore the vertex will be a\r\n" );
document.write( "maximum.\r\n" );
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document.write( "The vertex of a parabola is given by \r\n" );
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document.write( "x-coordinate of the vertex = \r\n" );
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document.write( "y-coordinate of the vertex = what you get when you substitute\r\n" );
document.write( "                             the x-coordinate in the equation\r\n" );
document.write( "                             and solve for y.\r\n" );
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document.write( "So for your problem:\r\n" );
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document.write( "x-coordinate of the vertex =  =  = 20\r\n" );
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document.write( "y-coordinate of the vertex = y = (20)² + 20(20) + 0 = 200\r\n" );
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document.write( "So the maximum area that such a right triangle can have is 200 \r\n" );
document.write( "and that will be when x = 20, which means that the other leg 40-x\r\n" );
document.write( "will also be 40-20 = 20.  This triangle will be an isosceles right\r\n" );
document.write( "triangle.  \r\n" );
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document.write( "Answer: Maximum area = 200cm² when both legs are 20cm each.\r\n" );
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document.write( "Edwin

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