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 intoa 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 foreach:

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)

Answer 1

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

Answer 2

summing the perimeters
3x+2πr=3 m

r=(3-3x)/2π

also

dr=(-3/2π)dx

now total area=(sqrt(3)/4 )*x² +πr²=A

dA/dx=(sqrt(3)/2 )x +2πrdr/dx

dA/dx=(sqrt(3)/2 )x +(3-3x)(-3/2π))

for maximum area

(sqrt(3)/2 )x +(3-3x)(-3/2π))=0

(sqrt(3)/2 )x +9(x-1)/2π))=0

solving for x

x=0.623 m

thus

the rope is cut at point (3*0.623=1.87 m ) for maximum area