document.write( "Question 1163357: Jula is constructing a pot to hold her precious Mathangias. The pot is square based rectangular prism. The surface area on the outside(not inside) is 500 cm2
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Algebra.Com's Answer #787439 by ikleyn(52790)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Let x be the size of the square base and h be the height of the pot, in centimeters.\r\n" );
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document.write( "Then the volume of the pot is\r\n" );
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document.write( "     V = \"x%5E2%2Ah\".                           (1)\r\n" );
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document.write( "The surface are of this pot, which has no top, is\r\n" );
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document.write( "      A = \"x%5E2+%2B+4xh\" = 500 cm^2.            (2)\r\n" );
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document.write( "So, we want to find optimal values of \"x\" and \"h\" to maximize the volume (1) at given restriction (2) on surface area.\r\n" );
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document.write( "From (2), we have \r\n" );
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document.write( "     h = \"%28500+-+x%5E2%29%2F%284x%29\" = \"125%2Fx+-+x%2F4\".            (3)\r\n" );
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document.write( "Substitute it into (1) to get\r\n" );
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document.write( "     V = \"x%5E2%2A%28125%2Fx+-+x%2F4%29%29\" = \"125x\" - \"x%5E3%2F4\".     (4)\r\n" );
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document.write( "Now we need to find a maximum value for V in formula (4) considering the volume  as a function of \"x\" only.\r\n" );
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document.write( "For it, take the derivative of  V(x) and equate it to zero\r\n" );
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document.write( "     V'(x) = 125 - \"%283x%5E2%29%2F4\" = 0.\r\n" );
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document.write( "It implies\r\n" );
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document.write( "     500 = 3x^2\r\n" );
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document.write( "     x^2 = \"500%2F3\"\r\n" );
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document.write( "     x = \"sqrt%28500%2F3%29\" = 12.91 cm (approximately.\r\n" );
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document.write( "Then\r\n" );
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document.write( "     h = (see formula (3)) = \"125%2Fx+-+x%2F4\" = \"125%2F12.91+-+12.91%2F4\" = 6.45 cm.\r\n" );
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document.write( "So, the problem is just solved.\r\n" );
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document.write( "Optimal dimensions are: the square base size of 12.91 cm and the height of 6.45 cm.\r\n" );
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document.write( "The maximum volume is  \"V%5Bmax%5D\" = \"12.91%5E2%2A6.45\" = 1075 cm^3  (approximately).\r\n" );
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