document.write( "Question 1135912: An 8 inch wire is to be cut. One piece is to be bent into the shape of a square, whereas the other piece is to be bent into the shape of a rectangle whose length is twice the width. Find the width of the rectangle that will minimize the total area.
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Algebra.Com's Answer #753629 by ikleyn(52799) $\"\"$ $\"About$

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document.write( "Let x = the width of the rectangle (now unknown).\r\n" );
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document.write( "Then its length is 2x, according to the condition.\r\n" );
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document.write( "Then the perimeter of the rectangle is  x + 2x + x + 2x = 6x.\r\n" );
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document.write( "Then the perimeter of the square is the rest  8-6x  and its side is   = .\r\n" );
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document.write( "Then the total area is   = x*(2x) + .\r\n" );
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document.write( "Simplify this expression;  then take the derivative over x and equate it to zero.\r\n" );
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document.write( "In this way you will find x.\r\n" );
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document.write( "The rest is just technique.\r\n" );
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