A 30-in. piece of string is cut into two pieces. One piece is used to form a circle and the other to form a square. How shouls the string be cut so that the sum of the areas is a minimum? Round to the nearest tenth, if necessary.
Let x be the amount used to form the circle, and 30-x is used to form the square. The total enclosed area is
A = pi x^2 + (30-x)^2
= (pi+1)x^2 -60 x + 900
Differentiate that with respect t x and set the derivative equalo to zero.
2 (pi+1)x -60 = 0
x = 30/(1 + pi) inches
Check my work.
Thanks drwls. I came to the same conclusion after racking my brain over it for some time. I made the problem more difficult than it really was. Thanks again for responding.
I made two rather bad mistakes. The circular area that can be made piece of length x is
pi * [x/(2 pi)]^2 and the square area that can be enclosed with length (30-x) is
[1/4)(30-x)]^2 The total is
A = x^2/(4 pi) + [7.5 - (x/4)]^2
= x^2[1/(4 pi) + 1/16] -3.75x +7.5^2
x [(1/(2 pi) + 1/8] = 3.75
x[1/pi + 1/4] = 7.5
x(4/pi + 1) = 30
x = 30/[(4/pi) + 1]
Tha maximum area is formed when all of string is used to form a circle. This is a minimum area that can be formed.
Set dA/dx = 0
This used to be one of my favourite max/min type of questions when I taught this stuff a long time ago.
Here is a slightly different approach from the solution done by drwls.
let the radius of the circle be r,
then the side of the square is (30-2(pi)r)/4 = (15-(pi)r)/2
A = [(15-(pi)r)/2]^2 + (pi)r^2
finding A' and setting this equal to zero gives r = 15/(pi+4)
then 2(pi)r, the length needed for the circle is 2(pi)(15/(pi+4)) = 30pi/(pi+4) = 13.2 , the same result as drwls
To find the cut that minimizes the sum of the areas of the circle and the square, we need to consider the formulas for the area of each shape.
Let's consider the circle first. The formula for the circumference (C) of a circle is given by C = 2πr, where r is the radius. In this case, the string is used to form the circumference, so we have 2πr = 30. From this equation, we can solve for r as follows:
2πr = 30
r = 30 / (2π)
r ≈ 4.77 (rounded to the nearest hundredth)
So the radius of the circle should be approximately 4.77 inches.
Next, let's consider the square. The formula for the perimeter (P) of a square is given by P = 4s, where s is the length of each side. In this case, the remaining piece of string is used to form the perimeter, so we have 4s = 30. From this equation, we can solve for s as follows:
4s = 30
s = 30 / 4
s = 7.5
So each side of the square should be 7.5 inches.
Finally, we can calculate the areas of the circle and the square. The area of a circle (A) is given by A = πr^2, and the area of a square (A) is given by A = s^2. We can plug in the values we found before:
Area of the circle = π(4.77)^2 ≈ 71.63 square inches (rounded to the nearest hundredth)
Area of the square = (7.5)^2 = 56.25 square inches
The sum of the areas is approximately 71.63 + 56.25 = 127.88 square inches.
Therefore, to minimize the sum of the areas of the circle and the square, the string should be cut such that the radius of the circle is approximately 4.77 inches, and each side of the square is 7.5 inches.