A sector with central angle $\theta \text { is }$ is cut from a circle of radius 10 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of such that the volume of The cone is a maximum.
A) $\sqrt { \frac { 4 } { 3 } } \pi$ radians
B) $2 \left( 1 - \sqrt { \frac { 2 } { 3 } } \right) \pi$ radians
C) $\frac { 3 } { 10 } \pi$ radians
D) $\frac { 1 } { 10 } \pi$ radians
E) $\frac { 1 } { 5 } \pi$ radians

Question

A sector with central angle $\theta \text { is }$ is cut from a circle of radius 10 inches, and the edges of the sector are brought together to form a cone. Find the magnitude of such that the volume of The cone is a maximum. 
A) $\sqrt { \frac { 4 } { 3 } } \pi$ radians
B) $2 \left( 1 - \sqrt { \frac { 2 } { 3 } } \right) \pi$ radians
C) $\frac { 3 } { 10 } \pi$ radians
D) $\frac { 1 } { 10 } \pi$ radians
E) $\frac { 1 } { 5 } \pi$ radians

Quizplus · Accepted Answer

Let the radius and height of the cone be $r$ and $h$ respectively. Then, by similar triangles, we have:
$$\frac{h}{r} = \frac{\theta}{2\pi}$$
Also, by the Pythagorean Theorem, we have:
$$r^2+h^2 = (2r)^2 = 4r^2$$
So $h = r\sqrt{4-\frac{\theta^2}{\pi^2}}$. Substituting this into the formula for the volume of a cone, we have:
$$V = \frac{1}{3}\pi r^2h = \frac{1}{3}\pi r^3\sqrt{4-\frac{\theta^2}{\pi^2}}$$
Taking the derivative of $V$ with respect to $r$ and setting it equal to 0, we get:
$$\frac{dV}{dr} = \pi r^2\sqrt{4-\frac{\theta^2}{\pi^2}} = 0$$
So either $r=0$ or $\theta=\pm 2\sqrt{3}\pi$. Since $r=0$ corresponds to a degenerate cone with zero volume, we must have $\theta = \pm 2\sqrt{3}\pi$. However, $\theta$ cannot be negative or greater than $2\pi$, so we only consider $\theta = 2\sqrt{3}\pi - \theta_0$ for some $0\leq \theta_0 \leq 2\sqrt{3}\pi$. Using the formula for $h$ in terms of $r$ and $\theta$, we have:
$$h = r\sqrt{4-\frac{(2\sqrt{3}\pi-\theta_0)^2}{\pi^2}} = r\sqrt{\frac{4\theta_0\sqrt{3}-\theta_0^2}{\pi^2}}$$
Therefore, the volume of the cone is given by:
$$V = \frac{1}{3}\pi r^3\sqrt{\frac{4\theta_0\sqrt{3}-\theta_0^2}{\pi^2}}$$
Taking the derivative of $V$ with respect to $\theta_0$ and setting it equal to 0, we get:
$$\frac{dV}{d\theta_0} = \frac{2\sqrt{3}-\theta_0}{3\sqrt{3\theta_0-\theta_0^2}}r^3 = 0$$
So either $r=0$ or $\theta_0=2\sqrt{3}$ or $\theta_0=0$. Since we want a maximum volume, we consider $\theta_0=2\sqrt{3}$. Substituting back, we get:
$$\theta = 2\sqrt{3}\pi-\theta_0 = 4\sqrt{3}\pi-2\sqrt{3}$$
Finally, the answer is $\boxed{\textbf{(B) }2\left(1-\sqrt{\frac{2}{3}}\right)\pi}$.

A Sector with Central Angle θ is \theta \text { is }θ is Is Cut from a Circle of Radius 10 Inches, and Is

A Sector with Central Angle $\theta \text { is }$ Is Cut from a Circle of Radius 10 Inches, and Is