document.write( "Question 388350: What is the smallest perimeter that can be used to make a rectangle with an area of 256cm squared? \n" ); document.write( "

Algebra.Com's Answer #274662 by stanbon(75887) $\"\"$ $\"About$

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What is the smallest perimeter that can be used to make a rectangle with an area of 256cm squared?
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\n" ); document.write( "Area = LW
\n" ); document.write( "256 = LW
\n" ); document.write( "W = (256/L)
\n" ); document.write( "-------
\n" ); document.write( "Perimeter = 2(L+W)
\n" ); document.write( "P = 2(L + (256/L))
\n" ); document.write( "---
\n" ); document.write( "P = 2(L^2+256)/L
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\n" ); document.write( "Take the derivative to get:
\n" ); document.write( "P' = 2[(L^2+256)-L(2L)]/L^2
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\n" ); document.write( "Solve: 2[(L^2+256)-L(2L)]/L^2 = 0
\n" ); document.write( "---
\n" ); document.write( "Solve: L^2+256-2L^2 = 0
\n" ); document.write( "-L^2 = -256
\n" ); document.write( "L^2 = 256
\n" ); document.write( "---
\n" ); document.write( "L = sqrt(256) = 16
\n" ); document.write( "Therefore W = sqrt(256) = 16
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\n" ); document.write( "Perimeter = 4*16 = 64
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\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H.
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