You have a wire that is 100 cm long. You wish to cut it into two pieces. One piece will be bent into the shape of a right triangle with legs of equal length. The other piece will be bent into the shape of a circle. Let A represent the total area enclosed by the triangle and the circle. What is the circumference of the circle when A is a minimum?
let each of the equal sides of the triangle be x
let the radius of the circle be r
area = (1/2)(x)(x) + πr^2
= x^2/2 + πr^2
we know that if the hypotenuse of the triangle is is h
x^2 + x^2 = h^2
2x^2 = h^2
h = x√2
then 2x + √2 x + 2πr = 100
r = (100 - 2x - √2x)/(2π)
then area
= x^2 /2 + π(100 - 2x - √2x)^2 /(4π^2)
simplify, differentiate, set the derivative equal to zero and solve for x
Then find r and the circumference of the circle
Ac = c^2 / 4 π
triangle perimeter = Pt = 100 - c
At = [Pt / (2 + √2)]^2 / 2
... = (Pt)^2 / (12 + 8√2)
... = (1E4 - 200 c + c^2) / (12 + 8√2)
A = (c^2 / 4π) +
... [(1E4 - 200 c + c^2) / (12 + 8√2)]
the minimum A lies on the axis of symmetry ... c = -b / 2a
c = [50 / (3 + 2√2)] / {(1 / 4π) + [1 / (12 + 8√2)]}
To find the circumference of the circle when the total area, A, is a minimum, we need to optimize the area function with respect to the length of the wire used for the circle.
Let's break down the problem step by step:
1. Let x represent the length of the wire used for the triangle. Since the triangle has two equal legs, each leg will have a length of x/2.
2. The remaining wire used for the circle will be (100 - x) cm.
3. The area of the triangle can be calculated using the formula: A_triangle = (1/2) * base * height. In this case, the base and height are both x/2, so the area of the triangle is A_triangle = (1/2) * (x/2) * (x/2) = x^2/8.
4. The circumference of the circle, C_circle, is given by the formula: C_circle = 2πr, where r is the radius of the circle. We need to find the value of r that minimizes the area A.
5. The circumference of the circle is equal to the remaining wire length, so C_circle = 100 - x.
6. To minimize the area A = A_triangle + A_circle, we need to find the value of x that makes the derivative of A with respect to x equal to zero.
7. The area of the circle, A_circle, can be calculated using the formula: A_circle = πr^2. Since r = (100 - x) / (2π), we can substitute this expression into the equation, which gives A_circle = π(100 - x)^2 / (4π^2).
8. Now, we can write the total area A as a function of x: A(x) = A_triangle + A_circle = x^2 / 8 + π(100 - x)^2 / (4π^2).
9. Taking the derivative of A(x) with respect to x, we have A'(x) = (2x) / 8 + (2π(100 - x)(-1)) / (4π^2).
10. Simplifying A'(x), we get A'(x) = x / 4π + (100 - x) / (2π^2).
11. To find the value of x that minimizes A, we set A'(x) = 0 and solve for x: x / 4π + (100 - x) / (2π^2) = 0.
12. Simplifying the equation, we have x + (100 - x) / (2π) = 0.
13. Multiplying through by 2π, we get 2πx + 100 - x = 0.
14. Rearranging the equation, we have 2πx = x - 100.
15. Dividing both sides by x, we get 2π = 1 - 100/x.
16. Rearranging again, we have 100/x = 1 - 2π.
17. Solving for x, we find x = 100 / (2π - 1).
18. Now that we have the value of x, we can calculate the circumference of the circle: C_circle = 100 - x.
19. Substituting the value of x, we get C_circle = 100 - (100 / (2π - 1)).
So, the circumference of the circle when the total area A is a minimum is 100 - (100 / (2π - 1)) cm.