SOLUTION: 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 ar
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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 Answer by richard1234(7193) (Show Source):
You can put this solution on YOUR website! 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.
Without loss of generality, let the perimeter of the square be L. Then, the side length is L/4 and the area is .
If the perimeter of the triangle is 10-L, then each side length is , and the height is . It follows that the area is .
Therefore, the total area is . 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.
