A wire 3 meters long is cut into two pieces. One piece is bent into a equilatera

Question

A wire 3 meters long is cut into two pieces. One piece is bent into a equilateral triangle for a frame for a stained glass ornament, while the other piece is bent into a circle for a TV antenna. To reduce storage space, where should the wire be cut to minimize the total area of both figures? Give the length of wire used for each:

Where should the wire be cut to maximize the total area? Again, give the length of wire used for each:

For the equilateral triangle:

For the circle:

(for both, include units)

Explanation / Answer

x = radius of circle y = side of triangle Total length = 3y + 2px = 3 total area = y²v3/4 + px² 3y + 2px = 3 y = (3–2px)/3 A = y²v3/4 + px² A = [(3–2px)/3]²v3/4 + px² A = v3/36(3–2px)² + px² A = v3/9(4–px)² + px² A = v3/9(16–3px+p²x²) + px² A = (16/9)v3 – (3/9)v3px + (v3/9)p²x² + px² A = (v3/9+1)p²x² – (3/9)v3px + (16/9)v3 differentiate, set equal to zero and solve for x A' = 2(v3/9+1)p²x – (3/9)v3p = 0 2(v3/9+1)px = (3/9)v3 x = (3/9)v3 / 2(v3/9+1)p = 1.5396 / 7.492 = 0.524 Aprox

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