document.write( "Question 596257: FInd the area of the largest rectangle that can be inscribed in a semicircle of fadius r. \n" ); document.write( "
Algebra.Com's Answer #377658 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Let x represents half the length of the\r\n" );
document.write( "rectangle, the length of rectangle = 2x\r\n" );
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document.write( "Let y represent the height of the rectangle. \r\n" );
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document.write( "Then the Area of the rectangle is\r\n" );
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document.write( "Area = length × width\r\n" );
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document.write( "   A = (2x)y\r\n" );
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document.write( "   A = 2xy\r\n" );
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document.write( "However we must now express y in terms of x and r.\r\n" );
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document.write( "Draw in a radius (which equals r) from the \r\n" );
document.write( "center of the semicircle to the upper right\r\n" );
document.write( "corner of the rectangle:\r\n" );
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document.write( "Use the Pythagorean theorem on the right triangle:\r\n" );
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document.write( "x² + y² = r²\r\n" );
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document.write( "     y² = r² - x²\r\n" );
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document.write( "      y = Ör² - x²\r\n" );
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document.write( "So substitute this for y in\r\n" );
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document.write( "      A = 2xy\r\n" );
document.write( "             _______\r\n" );
document.write( "      A = 2xÖr² - x²\r\n" );
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document.write( "We could take the derivative in this form, but it'll be \r\n" );
document.write( "easier if we square both sides first to get rid of the \r\n" );
document.write( "square root:\r\n" );
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document.write( "     A² = 4x²(r² - x²)\r\n" );
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document.write( "     A² = 4r²x² - 4x4\r\n" );
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document.write( "Remember that r is a constant.\r\n" );
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document.write( " 2A\"dA%2F%28dx%29\" = 8r²x - 16x³\r\n" );
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document.write( "Solve for \"dA%2F%28dx%29\"\r\n" );
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document.write( "   \"dA%2F%28dx%29\" = \"%288r%5E2x+-+16x%5E3%29%2FA\"\r\n" );
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document.write( "We set that equal to 0 to find the extremum:\r\n" );
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document.write( "    \"%288r%5E2x+-+16x%5E3%29%2FA\" = 0\r\n" );
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document.write( "Multiply both sides by A\r\n" );
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document.write( "          8r²x - 16x³ = 0\r\n" );
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document.write( "         8x(r² - 2x²) = 0\r\n" );
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document.write( "     8x = 0;  r² - 2x² = 0\r\n" );
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document.write( "      x = 0;        r² = 2x²\r\n" );
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document.write( "Obviously the minimum area is when x = 0\r\n" );
document.write( "                 only take positive square roots for equation on the right:\r\n" );
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document.write( "                     r = Ö2x²\r\n" );
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document.write( "                     r = xÖ2\r\n" );
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document.write( "                     \"r%2Fsqrt%282%29\" = x\r\n" );
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document.write( "We find y from:\r\n" );
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document.write( "                         _______  \r\n" );
document.write( "                    y = Ör² - x²\r\n" );
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document.write( "We need to substitute x² = \"r%5E2%2F2\"\r\n" );
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document.write( "                    y = \"sqrt%28r%5E2+-+r%5E2%2F2%29\"\r\n" );
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document.write( "                    y = \"sqrt%282r%5E2%2F2+-+r%5E2%2F2%29\"\r\n" );
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document.write( "                    y = \"sqrt%28r%5E2%2F2%29\"\r\n" );
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document.write( "                    y = \"r%2Fsqrt%282%29\"\r\n" );
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document.write( "So the largest rectangle has base 2x = \"2r%2Fsqrt%282%29\"\r\n" );
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document.write( "and height y = \"r%2Fsqrt%282%29\"\r\n" );
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document.write( "To find the maximum area, substitute in\r\n" );
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document.write( "                    A = (2x)y\r\n" );
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document.write( "                    A = \"2r%2Fsqrt%282%29\"\"%22%22%2A%22%22\"\"r%2Fsqrt%282%29\"\r\n" );
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document.write( "                    A = \"2r%5E2%2F2\"\r\n" );
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document.write( "                    A = r²\r\n" );
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document.write( "Edwin
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