document.write( "Question 962250: An open rectangular box with square base and open top is to contain 1000cm^3.Find the dimensions that require the least amount of material.Neglect the thickness of the material and waste in construction.(Hint:Here we are looking at surface area)\r
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Algebra.Com's Answer #587969 by ankor@dixie-net.com(22740)\"\" \"About 
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An open rectangular box with square base and open top is to contain 1000cm^3.
\n" ); document.write( "Find the dimensions that require the least amount of material.
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\n" ); document.write( "let x = side of the square base
\n" ); document.write( "let h = the height of the box
\n" ); document.write( "then the volume
\n" ); document.write( "x * x * h = 1000
\n" ); document.write( "x^2h = 1000
\n" ); document.write( "h = \"1000%2Fx%5E2\"
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\n" ); document.write( "The surface area of an open box
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\n" ); document.write( "S.A. = bottom area + 4 side areas
\n" ); document.write( "S.A. = x^2 + 4(x*h)
\n" ); document.write( "Replace h with \"1000%2Fx%5E2\"}
\n" ); document.write( "S.A. = x^2 + 4(x*\"1000%2Fx%5E2\")
\n" ); document.write( "cancel x into x^2
\n" ); document.write( "S.A. = x^2 + \"4000%2Fx\"
\n" ); document.write( "Graph this in your graphing calc S.A. = y
\n" ); document.write( "\"+graph%28+300%2C+200%2C+-5%2C+20%2C+-200%2C+1000%2C+x%5E2%2B%284000%2Fx%29%29+\"
\n" ); document.write( "minimum surface when x = 12.6 cm the side of the square base
\n" ); document.write( "Find the height
\n" ); document.write( "h = \"1000%2F12.6%5E2\"
\n" ); document.write( "h = \"1000%2F158.76\"
\n" ); document.write( "h = 6.3 cm is the height
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\n" ); document.write( "Summarize, 12.6 by 12.6 by 6.3 dimensions for minimum surface area
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\n" ); document.write( "confirm this by finding the volume with these dimension
\n" ); document.write( "12.6 * 12.6 * 6.3 = 1000.2, close enough
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