document.write( "Question 395574: A carpenter is building a rectangular room with a fixed perimeter of 304 ft. What dimensions would yield the maximum area? What is the maximum area? \r
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Algebra.Com's Answer #280846 by richard1234(7193) $\"\"$ $\"About$

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There are multiple ways to prove that the maximum area is a square. You could let A = w*(152 - w), then take the derivative or find the vertex, as the other tutor did.\r
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\n" ); document.write( "\n" ); document.write( "I like the classic AM-GM inequality: AM-GM (arithmetic mean - geometric mean inequality) says that for positive real numbers $\"a%5Bi%5D\"$ , the arithmetic mean of all $\"a%5Bi%5D\"$ 's is greater than or equal to the geometric mean. If we let w and 152-w be the sides,\r
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\n" ); document.write( "\n" ); document.write( " $\"%28w+%2B+%28152-w%29%29%2F2+%3E=+sqrt%28w%2A%28152-w%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"+76+%3E=+sqrt%28w%2A%28152-w%29%29\"$ \r
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\n" ); document.write( "\n" ); document.write( " $\"76%5E2+%3E=+w%2A%28152-w%29\"$ --> the area is less than or equal to 76^2. AM-GM says that the equality case occurs if and only if all the $\"a%5Bi%5D\"$ 's are equal, or w = 304-w --> w = 76, area = 76^2.\r
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\n" ); document.write( "\n" ); document.write( "This is a slightly unconventional way, but it's the easiest way to prove the similar problem that says that the largest possible volume of a rectangular solid of fixed surface area occurs when the solid is a cube. \n" ); document.write( "

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