document.write( "Question 1201504: An aquarium that holds 40 cubic meters of water is to be made such that the length of its base is twice the width. If material for the base costs $20 per square meter, and the material for the sides costs $16 per square meter, find the cost of the materials for the cheapest such aquarium. \n" ); document.write( "

Algebra.Com's Answer #835923 by ikleyn(52795) $\"\"$ $\"About$

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\n" ); document.write( "An aquarium that holds 40 cubic meters of water is to be made
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document.write( "Let  w  be the width of the aquarium;\r\n" );
document.write( "then its length is 2w, according to the problem.\r\n" );
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document.write( "If the height is h, then the volume is\r\n" );
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document.write( "    V = w*(2w)*h = 2w^2*h,\r\n" );
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document.write( "so\r\n" );
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document.write( "    2w^2*h = 40 cubic meters, or\r\n" );
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document.write( "     w^2*h = 20 cubic meters.\r\n" );
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document.write( "It gives  h = .      (1)\r\n" );
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document.write( "The base area is w*(2w) = 2w^2; the base cost is 20*2w^2 = 40w^2 dollars.\r\n" );
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document.write( "The lateral area is (w + 2w + w + 2w)*h = 6wh.  The lateral sides cost is 16*6wh = 96wh dollars.\r\n" );
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document.write( "The total cost is \r\n" );
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document.write( "    C = 40w^2 + 96wh = substitute h from (1) =  =  +  =  + .\r\n" );
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document.write( "So, we want to minimize this function C(w) =  + .    (2)\r\n" );
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document.write( "To find the minimum, take the derivative and equate it to zero.\r\n" );
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document.write( "Doing it, you will get, step by step\r\n" );
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document.write( "    80w = \r\n" );
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document.write( "    80w^3 = 1920\r\n" );
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document.write( "      w^3 = 1920/80 = 24\r\n" );
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document.write( "    w =  = 2.88.\r\n" );
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document.write( "Thus the width is 2.885 m;  the length is twice of it 2*2.885 = 5.77 m.\r\n" );
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document.write( "                            the height is   =  = 2.41 m.\r\n" );
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document.write( "The cheapest cost is  C = formula (2) =  +  = 998.43 dollars.\r\n" );
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