document.write( "Question 142358This question is from textbook McDougal littell Math
\n" ); document.write( ": What is the largest possible area of a rectangle with integer side length and a perimeter of 22? \n" ); document.write( "
Algebra.Com's Answer #103652 by solver91311(24713)\"\" \"About 
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\"P=2L+%2B+2W\"
\n" ); document.write( "\"A=LW\"\r
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\n" ); document.write( "\n" ); document.write( "Given P = 22, so \"2L+%2B+2W=22\"\r
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\n" ); document.write( "\n" ); document.write( "A little algebra gets us: \"L=11-W\"\r
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\n" ); document.write( "\n" ); document.write( "So \"A=LW=%2811-W%29W=-W%5E2%2B11W\"\r
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\n" ); document.write( "\n" ); document.write( "This equation graphs to a convex down parabola with vertex at \"-11%2F%282%28-1%29%29=11%2F2\", telling us that the maximum area is achieved when the length of the side is \"11%2F2\". The difficulty is that the problem asks for the maximum area of a rectangle with integer side lengths. The nearest integer smaller than \"11%2F2\" is 5 and the nearest integer larger than \"11%2F2\" is 6.\r
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\n" ); document.write( "\n" ); document.write( "Substituting 5 for W in \"2L+%2B+2W=22\" gives us \"L=6\", so obviously substituting 6 for W in \"2L+%2B+2W=22\" will give us \"L=5\". Since 5 and 6 are the nearest integers to the side length that provides the maximum area (notice that all 4 sides would be \"11%2F2\" making a square) these must be the required side lengths and the maximum area for integer side lengths is \"5%2A6=30\"
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