•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 enclosed by the triangle and the circle (A) is minimum, we need to express the area of both shapes in terms of a single variable.
Let's assume that each leg of the right triangle formed is l cm. Then, the hypotenuse of the right triangle will be sqrt(2) times the length of each leg. Therefore, the perimeter of the right triangle will be:
Perimeter of right triangle = 2l + sqrt(2)l + sqrt(2)l
= (2 + 2sqrt(2))l
The remaining wire, which forms the circle, will have a length of:
Wire for the circle = 71 - (2 + 2sqrt(2))l
Now, let's find the radius (r) of the circle. The circumference (C) of a circle is given by the formula C = 2πr.
Since the remaining wire is used to form the circle, we can write:
2πr = 71 - (2 + 2sqrt(2))l
r = (71 - (2 + 2sqrt(2))l) / (2π)
The area of the circle (A_circle) is given by the formula A_circle = πr^2:
A_circle = π((71 - (2 + 2sqrt(2))l) / (2π))^2
= (71 - (2 + 2sqrt(2))l)^2 / (4π)
Now let's express the area of the right triangle (A_triangle). The area of a right triangle is given by the formula A_triangle = (1/2) * base * height, where the base and height are the legs of the triangle:
A_triangle = (1/2) * l * l
= l^2 / 2
Finally, the total area (A) is the sum of the areas of the circle and the triangle:
A = A_circle + A_triangle
= (71 - (2 + 2sqrt(2))l)^2 / (4π) + l^2 / 2
To find the circumference of the circle when A is a minimum, we need to differentiate A with respect to l, set the derivative equal to zero, and solve for l. Let's do that:
dA / dl = (2(71 - (2 + 2sqrt(2))l)(-2 - 2sqrt(2)) / (4π)) + (2l / 2)
= - ((2 + 2sqrt(2))(2 - 71 + (2 + 2sqrt(2))l) / π) + l
Setting dA / dl = 0 and solving for l:
- ((2 + 2sqrt(2))(2 - 71 + (2 + 2sqrt(2))l) / π) + l = 0
(2 + 2sqrt(2))(2 - 71 + (2 + 2sqrt(2))l) = πl
Simplifying the equation:
4 - 142 + (8 + 8sqrt(2))l + (4 + 4sqrt(2))l - 142sqrt(2) - 142sqrt(2)l = πl
-284 + (12 + 12sqrt(2))l - 142sqrt(2) - 142sqrt(2)l = πl
-284 - 142sqrt(2) = πl - (12 + 12sqrt(2) + 142sqrt(2) + 142sqrt(2))l
-284 - 142sqrt(2) = πl - (12 + 298sqrt(2))l
-284 - 142sqrt(2) = πl - 310l - 298πsqrt(2)l
-284 - 142sqrt(2) = (π - 310 - 298πsqrt(2))l
l = -284 - 142sqrt(2) / (π - 310 - 298πsqrt(2))
Once we have the value of l, we can substitute it back into the equation for the radius of the circle (r = (71 - (2 + 2sqrt(2))l) / (2π)) to find the radius, and then calculate the circumference (C = 2πr) of the circle at the minimum area, A.
To solve this problem, we need to find the dimensions of the triangle and the circle that minimize the total area, A. Let's start with the triangle.
Let's assume that the legs of the right triangle have length 'a', which means that the hypotenuse is also 'a' due to the equal legs. We can use the Pythagorean theorem to relate the sides of the triangle:
a^2 = a^2 + a^2
a^2 = 2a^2
a = sqrt(2)
Now we know the dimensions of the triangle. The next step is to determine the radius of the circle.
Let's represent the radius of the circle as 'r'. We know that the perimeter of the circle is equal to the length of the remaining wire after cutting it for the triangle, which is 71 cm minus the perimeter of the triangle.
Perimeter of the triangle = hypotenuse + leg1 + leg2
Perimeter of the triangle = a + a + a
Perimeter of the triangle = 3a
Length of the remaining wire = 71 cm - 3a
Since the remaining wire is used to form the circle, its circumference should be equal to the length of the remaining wire:
2πr = 71 cm - 3a
Now we have an equation involving the radius 'r' and the side length 'a' of the right triangle.
Next, we express the area, A, in terms of 'r' and 'a':
A = (area of triangle) + (area of circle)
A = (1/2) * a * a + π * r^2
Now we can substitute the value of 'a' in terms of 'r' in the equation for the area:
A = (1/2) * (sqrt(2))^2 + π * r^2
A = 1 + πr^2
Now we have the total area, A, only in terms of 'r'. To find the minimum value of A, we can take the derivative of A with respect to 'r' and set it to zero:
dA/dr = 0
2πr = 0
r = 0
Since the radius of the circle cannot be zero, we need to consider the endpoints. The radius 'r' must be positive, so it can't be the endpoint. Hence, the other endpoint must be the minimum value. So, the minimum value of A occurs when the remaining wire forms a semi-circle, when r = (71 - 3a) / (2π).
Now we can substitute the value of 'a' we found earlier:
r = (71 - 3*sqrt(2)) / (2π)
Therefore, the circumference of the circle when the area is a minimum is given by:
Circumference = 2πr
Circumference = 2π * ((71 - 3*sqrt(2)) / (2π))
Circumference = 71 - 3*sqrt(2)
Thus, the circumference of the circle when the area is a minimum is 71 - 3*sqrt(2) cm.
You can assign x for the radius of the circle, or as one of the equal legs of the right triangle, but either way, the answer above is incorrect and unhelpful. To go from x = equal length triangle leg use this total area formula to graph a parabola, then find the minimum value for x which will be your triangle leg length, then find the perimeter of the triangle first, then circle from there.
1/2x^2 + (71-(2x + sqrt2x)/2pi)^2
3s+c = 71
a = √3/4s^2 + c = √3/4 s^2 -3s + 71
da/ds = √3/2 s - 3
clearly a is minimum when s = 2√3
so, the circumference c is 71-6√3