
calculus min?
A wire 7 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:
For the equilateral triangle:
For the circle:
(for both, include units)
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
Let the circumference of the circle be x. Then the perimeter of the triangle is (7-x).
The diameter of the circle is x/π, and its radius is x/(2π). Thus, its area is:
Ac = π(x/2π)² = x²/(4π)
One side of the triangle is (7-x)/3. Its height is √3/2 times as side. Thus, its area is:
At = (1/2)(7 - x)(√3/2)(7 - x)/3
At = (√3/12)(49 - 14x + x²)
Now, to minimize total area, take the derivative and set to 0:
A = Ac + At
A' = x/(2π) + (√3/12)(2x - 14)
A' = x/(2π) + x√3/6 - 7√3/6
0 = x(√3/6 + 1/(2π)) - 7√3/6
x = (7√3/6) / (√3/6 + 1/(2π))
x ≈ 4.512
That makes the length of the triangle piece 7-x = 2.488
To maximize area, it must be at one of the endpoints, x=0 or x=7. Check both:
x=0: A = (1/2)(7)(√3/2)(7)/3 ≈ 7.073
x=7: A = 7²/(4π) ≈ 3.899
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