Calculas H/W help

Question

You have a wire that is 83 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a square. The other piece will be bent into the shape of a circle. Let A represent the total area of the square and the circle. What is the circumference of the circle when A is a minimum?

Answer 1

4s = the square's perimeter = 83 - C

s = (83-C)/4

s^2 = area of the square = [(83-C)/4]^2 = (6889 -166C +C^2)/16 = 6889/16 - 166C/16 + C^2/16

pi(r^2) = area of the circle

2pi(r) = circumference of the circle = C

r=C/2pi

Area of the circle = pi(C/2pi)^2 = C^2/4pi

total area = C^2/4pi + 430.5625 - 10.375C + 0.0625C^2

= (1/4pi + 1/16)C^2 - 10.375C + 6889/16

= about (.14)C^2 -10.4C + 430.6

-b/2a = 10.4/.28 = 37.1 is the x coordinate of the parabola's vertex, the Circumference which minimizes total area: 37.1 cm

C=37.1 = 2pi(r)

r = 37.1/2pi = 37.1/6.28 = 5.9

area of the circle = pi(r^2) = pi(5.9^2) = 109.6

perimeter of the square = 83 -37.1 = 45.9 cm

square's side = 45.9/4 = 11.5

total area =11.5^2 + 109.6 = 132.25 + 109.6 = 241.85 cm^2

smallest total = .14(37.1)^2 -10.4(37.1) + 430.6 = 192.7 - 385.8 + 430.6 = 237.5

A'(C) = .28C -10.4 = 0

C=10.4/.28 = 37.1 cm

rounding errors and approximating pi cause a slight discrepancy,but minimum total area is about 240 cm^2

but squares always have areas larger than circles with the same perimeter/circumference

45.9>37.1, giving some confidence in the result, the wire has to be cut giving the square a greater area to minimize total area.

Answer 2

P = 83 - x

s = (83 - x) / 4

A = (83 - x)² / 16

C = x

C = 2πr

x / 2π = r

A = πr²

A = π(x / 2π)²

A = x / 4π

A_total = [(83 - x)² / 16} + [x / 4π]

Determine first derivative and solve for 0.

Answer 3

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