document.write( "Question 310349: The base of a rectangle is on the x-axis and its two upper vertices are on the parabola y=16-x^2. of all such rectangles, what are the dimensions of the one with greatest area? \n" ); document.write( "
Algebra.Com's Answer #221899 by Edwin McCravy(20056)\"\" \"About 
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document.write( "I think you need to draw a graph to understand what is going on:\r\n" );
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document.write( "The base of the green rectangle is 2x, because\r\n" );
document.write( "the base extends from -x to +x on the x axis,\r\n" );
document.write( "and that is a distance of 2x.\r\n" );
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document.write( "The height of the green rectangle is y, so\r\n" );
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document.write( "The area of the green rectangle is\r\n" );
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document.write( "Area = base * height or\r\n" );
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document.write( "\"A+=+%282x%29%2Ay\"\r\n" );
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document.write( "Since y and x are related by the equation of the parabola,\r\n" );
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document.write( "we can substitute \"%2816-x%5E2%29\" for \"y\"\r\n" );
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document.write( "\"A+=+%282x%29%2A%2816-x%5E2%29\"\r\n" );
document.write( "\"A+=+32x-2x%5E3\"\r\n" );
document.write( "\"%28dA%29%2F%28dx%29=32-6x%5E2\"\r\n" );
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document.write( "Set \"%28dA%29%2F%28dx%29=32-6x%5E2=0\"\r\n" );
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document.write( "\"32-6x%5E2=0\"\r\n" );
document.write( "\"16-3x%5E2=0\"\r\n" );
document.write( "\"16=3x%5E2\"\r\n" );
document.write( "\"16%2F3=x%5E2\"\r\n" );
document.write( "\"%22%22%2B-sqrt%2816%2F3%29=x\"\r\n" );
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document.write( "Use the positive value\r\n" );
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document.write( "Using the second derivative test,\r\n" );
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document.write( "\"%28dA%29%2F%28dx%29=32-6x%5E2\"\r\n" );
document.write( "\"%28d%5E2A%29%2F%28dx%5E2%29=-12x\", which is negative when \"x=%284sqrt%283%29%29%2F3\",\r\n" );
document.write( "which proves that A is a maximum when \"x=%284sqrt%283%29%29%2F3\".\r\n" );
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document.write( "Therefore, the base of the largest rectangle\"%22%22=2x=2%2A%28%284sqrt%283%29%29%2F3%29=%288sqrt%283%29%29%2F3\", \r\n" );
document.write( "and its height is\"%22%22=y+=+16-x%5E2=16-16%2F3=48%2F3-16%2F3=32%2F3\".\r\n" );
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document.write( "Edwin

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