document.write( "Question 389682: Prove or disprove: The rectangular solid of minimum surface area with fixed volume is a cube. \n" ); document.write( "
Algebra.Com's Answer #276214 by richard1234(7193)\"\" \"About 
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Without loss of generality, let the fixed volume be 1, and the dimensions of the rectangular solid be x, y, z. Then,\r
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\n" ); document.write( "\n" ); document.write( "\"xyz\" = 1, and we want to show whether\r
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\n" ); document.write( "\n" ); document.write( "\"2%28xy+%2B+yz+%2B+zx%29+%3E=+6\" --> \"xy+%2B+yz+%2B+zx+%3E=+3\" is true (if it is true, then the solid must have a surface area of at least 6, and the solid of least surface area is a cube).\r
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\n" ); document.write( "\n" ); document.write( "Applying the AM-GM inequality,\r
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\n" ); document.write( "\n" ); document.write( "\"%28xy+%2B+yz+%2B+zx%29%2F3+%3E=+root%283%2C+x%5E2y%5E2z%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "Since \"xyz+=+1\", replace it into the right-hand side and simplify:\r
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\n" ); document.write( "\n" ); document.write( "\"%28xy+%2B+yz+%2B+zx%29%2F3+%3E=+root%283%2C+1%29+=+1\"\r
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\n" ); document.write( "\n" ); document.write( "\"xy+%2B+yz+%2B+zx+%3E=+3\" as desired. The equality of AM-GM occurs when x = y = z, i.e. the rectangular solid is a cube.\r
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