document.write( "Question 379922: a piece of wire 10m long is cut into two pieces. one piece is bent into a square and the other is bent into an equilateral triangle. how should the wire be cut so that the total area enclosed is a)a maximum b) a minimum \n" ); document.write( "

Algebra.Com's Answer #269690 by richard1234(7193) $\"\"$ $\"About$

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Suppose the two wire lengths are L and 10-L. We need to derive the formulas for the areas of a square and equilateral triangle given the perimeter.\r
\n" ); document.write( "\n" ); document.write( "Without loss of generality, let the perimeter of the square be L. Then, the side length is L/4 and the area is $\"L%5E2%2F16\"$ .\r
\n" ); document.write( "\n" ); document.write( "If the perimeter of the triangle is 10-L, then each side length is $\"%2810-L%29%2F3\"$ , and the height is $\"%2810-L%29%28sqrt%283%29%29%2F6+\"$ . It follows that the area is

.\r
\n" ); document.write( "\n" ); document.write( "Therefore, the total area is $\"%28%284sqrt%283%29+%2B+9%29%2F144%29L%5E2+-+%285sqrt%283%29%2F9%29L+%2B+20sqrt%283%29%2F9\"$ . As the function is a second degree function, the vertex occurs at L = -b/2a = 4.3496. This represents the minimum total area, as the function points upward. As this function has no absolute maximum, the maximum total area must occur at an endpoint (either L = 0 or L = 10). Checking, we find that L=10 provides the optimal area. \n" ); document.write( "

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