You have a wire that is 71 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?
To find the circumference of the circle when the total area (A) is a minimum, we need to optimize the area function.
Let's denote the length of the legs of the right triangle as x. So, one leg of the right triangle will be x, and the other leg will also be x.
The remaining wire length after cutting x units will be (71 - 2x) cm, which will be used to make the circular shape.
The area of the right triangle can be calculated using the formula: A_triangle = (1/2) * base * height. In this case, both the base and height are x, so the area of the triangle becomes: A_triangle = (1/2) * x * x = (1/2) * x^2.
The circumference of the circle can be calculated using the formula: C_circle = 2 * π * r, where r is the radius of the circle.
To find the radius, we need to determine how much wire is left after forming the right triangle. Since each leg of the right triangle uses x units of wire, the remaining wire for the circle will be (71 - 2x) cm. We know the circumference of the circle is equal to the length of the wire, so we can write:
2 * π * r = (71 - 2x)
Now, we need to express r in terms of x to substitute it into the area function. Divide both sides by 2π:
r = (71 - 2x) / (2π)
The area of the circle can be calculated using the formula: A_circle = π * r^2. Substituting the value of r:
A_circle = π * [(71 - 2x) / (2π)]^2 = (71 - 2x)^2 / (4π)
The total area, A, is given by A_total = A_triangle + A_circle:
A = (1/2) * x^2 + (71 - 2x)^2 / (4π)
To find the value of x that minimizes A, we can take the derivative of A with respect to x and set it equal to 0. Differentiating A with respect to x:
dA/dx = x - [(71 - 2x) * 2 / (4π)] = 0
Solving this equation for x will give us the length of the legs of the right triangle that minimizes the total area. Once we have the length of the legs, we can calculate the radius of the circle using the equation r = (71 - 2x) / (2π), and finally, substitute it into the circumference formula C_circle = 2 * π * r to find the circumference of the circle.
To find the circumference of the circle when the total area enclosed by the triangle and the circle, A, is a minimum, we need to first express the area A in terms of the circumference of the circle.
Let's start by defining the variables:
- Let x represent the length of one of the legs of the right triangle.
- Since the legs of the right triangle are equal in length, the other leg will also be x.
- The hypotenuse of the right triangle can be found using the Pythagorean theorem: hypotenuse = sqrt(x^2 + x^2) = sqrt(2x^2) = sqrt(2) * x.
The perimeter (P) of the right triangle is the sum of the lengths of its sides, which are x, x, and the hypotenuse (sqrt(2) * x): P = 2x + sqrt(2) * x.
Thus, the length of the remaining wire that will be used to make the circle is: 71 - (2x + sqrt(2) * x).
Now, let's express the area of the right triangle in terms of x:
- The area of a right triangle is given by A_triangle = (1/2) * base * height.
- In this case, the base and height of the triangle are both equal to x.
- Therefore, A_triangle = (1/2) * x * x = (1/2) * x^2.
The circumference (C) of a circle can be calculated using the formula: C = 2 * π * r, where r is the radius of the circle.
Since the remaining wire is used to create the circle, the circumference of the circle is: C = 2 * π * (remaining wire length / (2π)), which simplifies to C = remaining wire length.
To find the value of x that minimizes the total area A, we need to find the value of x that corresponds to the minimum value of A_triangle. Since x is a positive length, the minimum value of A_triangle occurs when x = 0, hence creating a degenerate triangle with an area of 0.
Therefore, to minimize the total area A, we need to use the entire wire to form a circle, making the length of the remaining wire the circumference of the circle. In this case, the circumference of the circle when A is a minimum is 71 cm.
This pops up in many forms. If the length of the wire is z, the side of the triangle is s, and the radius of the circle is r, then we have
3s + 2πr = z
so, s = (z-2πr)/3
The total area is
A = √3/4 s^2 + πr^2
= √3/36 (z-2πr)^2 + πr^2
You want A to be minumum, so dA/dr = 0
dA/dr = √3/36 * 2(z-2πr)(-2) + 2πr
= 2πr - π/√27 (z-2πr)
= (1+π/√27)(2πr) - 2πz/√27
dA/dr = 0 when r = z/(π+3√3) ≈ 8.51
The area of the circle is
πr^2 = π*(71/(π+3√3))^2 ≈ 227.81
double-check my math there.