You have a wire that is 26 cm long. You wish to cut it into two pieces.......... What is the circumference?

Question

You have a wire that is 26 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 minium?

Answer 1

Suppose that you dedicate x units of the wire for the circle and 26 - x units of the wire for the square. If so, the circumference of the circle is x and the perimeter of the square is 26 - x, so we have:

2πr = x and 4s = 26 - x ==> r = x/(2π) and s = (26 - x)/4.



Then, the area of the circle is πr^2 and the area of the square is s^2, so the sum of their areas is:

A = πr^2 + s^2

= π[x/(2π)]^2 + [(26 - x)/4]^2, since r = x/(2π) and s = (26 - x)/4

= x^2/(4π) + (1/16)x^2 - (13/4)x + 169/4

= [1/(4π) + 1/16]x^2 - (13/4)x + 169/4.



We now wish to minimize A. By differentiating A with respect to x:

dA/dx = 2[1/(4π) + 1/16]x - 13/4 = [1/(2π) + 1/8]x - 13/4;



this equals zero when:

[1/(2π) + 1/8]x - 13/4 = 0 ==> x = (13/4)/[1/(2π) + 1/8] = 26π/(π + 4).



Just a note: I multiplied the numerator and denominator of (13/4)/[1/(2π) + 1/8] by 8π, the LCD of the fractions in the numerator and denominator, to arrive at the simplified result.



Therefore, the circumference of the circle that minimizes A is:

C = x = 26π/(π + 4).



I hope this helps!

Answer 2

A(r) = pi r^2 + ((26 - 2 pi r) / 4)^2

A(r) = pi r^2 + ((13 - pi r) / 2)^2

A(r) = pi r^2 + (169 - 26 pi r + pi^2 r^2) / 4

A(r) = pi r^2 + 169/4 - 13/2 pi r + 1/4 pi^2 r^2

A'(r) = 2 pi r - 13/2 pi + 1/2 pi^2 r

A'(r) = 0



2 pi r - 13/2 pi + 1/2 pi^2 r = 0

(2 pi + 1/2 pi^2) r = 13/2 pi

r = ((13/2) pi) / ((2 + 1/2 pi) pi)

r = (13/2) / (2 + 1/2 pi)

r = 13 / (4 + pi)

2 pi r = 2 pi (13 / (4 + pi))

2 pi r = 26 pi / (pi + 4)

2 pi r = 11.4374 cm



So the area is a minimum when the circumference of the circle is about 11.4374 cm.



You can (to a reasonable extent) verify this by calculating the area when the circumference is 11.4 cm, 11.4374 cm, and 11.48 cm.



When c = 11.4 cm, r = 11.4 cm / 2pi = 1.8144 cm

When c = 11.4374 cm, r = 11.4374cm / 2pi = 1.8203 cm

When c = 11.48 cm, r = 11.48cm / 2pi = 1.8271 cm



A(r) = pi r^2 + ((26 - 2 pi r) / 4)^2

A(1.8144) = 23.664386 cm^2

A(1.8203) = 23.664189 cm^2

A((1.8271) = 23.664447 cm^2



Looks good! A(1.8203), which is the area associated with a circumference of 11.4374cm, is the smallest.

Answer 3

Say length of things of sq. is x, and radius of circle is r. Pi = P You get the equation 4x+2rP=80 from the sum of perimeter. Then area of two shapes is x^2+Pr^2. by skill of substituting x= (20-0.5Pr)^2 into 2nd equation you get (20-0.5Pr)^2+Pr^2= area. you may positioned it into your photos calculator to discover minimum or use calculus. you will get answer r=5.6 so circumference would be 35.19cm.