document.write( "Question 965124: Not sure if \"topic\": selection is correct.\r
\n" ); document.write( "\n" ); document.write( "how do you figure out the smallest and largest possible perimeters of an area:\r
\n" ); document.write( "\n" ); document.write( "i.e. (1) A picture frame has an area of 48 sq inches. What is the largest possible perimeter of the pic frame? (2) A backyard has an area of 64 sq feet. What is the smallest possible perimeter of the backyard?\r
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Algebra.Com's Answer #589868 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
Question 1 is a rectangular shape that can be analyzed. Question 2 shape is unspecified so is incomplete.\r
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\n" ); document.write( "\n" ); document.write( "x and y dimensions of the rectangle.
\n" ); document.write( "p for perimeter:
\n" ); document.write( "\"p=2x%2B2y\" and \"xy=48\"
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\n" ); document.write( "\"p=2x%2B2%2848%2Fx%29\"
\n" ); document.write( "\"p=2x%2B96%2Fx\"\r
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\n" ); document.write( "\n" ); document.write( "The only method for minimizing or maximizing the perimeter seems to be derivatives.\r
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\n" ); document.write( "\n" ); document.write( "\"dp%2Fdx=2%2B96%2A%28d%2Fdx%29%28x%5E%28-1%29%29\"
\n" ); document.write( "\"2%2B96%28-1%29%28x%5E-2%29\"
\n" ); document.write( "\"2-96%2Fx%5E2\"\r
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\n" ); document.write( "\n" ); document.write( "Finding extreme values, set derivative to zero. Continue other steps.
\n" ); document.write( "\"dp%2Fdx=2-96%2Fx%5E2=0\"
\n" ); document.write( "\"%282x%5E2-96%29%2Fx%5E2=0\"
\n" ); document.write( "Denominator cannot be zero, and for derivative be zero, the NUMERATOR must be equated to zero.\r
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\n" ); document.write( "\n" ); document.write( "\"2x%5E2-96=0\"
\n" ); document.write( "\"x%5E2-48=0\"
\n" ); document.write( "\"x%5E2=48\"
\n" ); document.write( "\"x=sqrt%2848%29\", the positive value only.\r
\n" ); document.write( "\n" ); document.write( "-
\n" ); document.write( "\"x=sqrt%282%2A24%29=sqrt%282%2A2%2A6%2A2%29=sqrt%282%2A2%2A2%2A2%2A3%29\"
\n" ); document.write( "\"highlight%28x=4%2Asqrt%283%29%29\"
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\n" ); document.write( "and from the original area equation, obviously \"y=4%2Asqrt%283%29\", making the area shape as a SQUARE shape. The values for MINIMUM perimeter.\r
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\n" ); document.write( "\n" ); document.write( "Note that you will find this is for MAXIMUM area. Check about the area equation and you should find no minimum.\r
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\n" ); document.write( "\n" ); document.write( "Here is a graph for the PERIMETER equation, in one or either variable, here being \"p=2x%2B48%2Fx\", which shows a MINIMUM for the perimeter:
\n" ); document.write( "\"graph%28350%2C350%2C-8%2C39%2C-8%2C39%2C2x%2B96%2Fx%29\"
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