Re: Max and min two
MarkFL said:
You really should be posting what you have tried so far, but I will get you started.
We are going to take a circular sheet of radius $R$, and remove from it a circular sector whose central angle is $\theta$, leaving a circular sector from which we are going to form a cone. The radius of this cone can be found by using the formula for the length of a circular arc:
$$r=R(2\pi-\theta)$$
To determine the height of the cone, we may observe that the area of the sector from which we are forming the cone is equal to the lateral surface area of the cone:
$$\pi r\sqrt{r^2+h^2}=\frac{1}{2}R^2(2\pi-\theta)$$
Solve this for $h$, and then state your objective function, which is the volume of the cone:
$$V=\frac{1}{3}\pi r^2h$$
Substitute into this the values for $h$ and $r$ so that you have a function of the variable $\theta$ and the constant $R$, and then maximize.
I wanted to repost as I originally made some errors which I want to correct.
We have:
$$r=\frac{R}{2\pi}(2\pi-\theta)$$
$$\pi r\sqrt{r^2+h^2}=\frac{1}{2}R^2(2\pi-\theta)=\pi rR$$
And so dividing through by $\pi r\ne0$ we obtain:
$$\sqrt{r^2+h^2}=R$$
Note: we could have more easily obtained this by simply analyzing the cross-section, but as they say hindsight is 20/20. :D
Squaring, we have:
$$r^2+h^2=R^2$$
Hence, isolating $h^2$ and substituting for $r$, we have:
$$h^2=R^2-\left(\frac{R}{2\pi}(2\pi-\theta) \right)^2=\left(\frac{R}{2\pi} \right)^2\theta(4\pi-\theta)$$
Taking the positive root, we obtain:
$$h=\frac{R}{2\pi}\sqrt{\theta(4\pi-\theta)}$$
And so now, the volume of the cone may be written as:
$$V=\frac{1}{3}\pi\left(\frac{R}{2\pi}(2\pi-\theta) \right)^2\left(\frac{R}{2\pi}\sqrt{\theta(4\pi-\theta)} \right)$$
$$V=\frac{R^3}{24\pi^2}(2\pi-\theta)^2\sqrt{\theta(4\pi-\theta)}$$
This is what you want to maximize, and after having worked out the problem, and looking at the answer you posted, the angle you are after is actually:
$$\beta=2\pi-\theta$$
So, in this case, you want to maximize:
$$V=\frac{R^3}{24\pi^2}\beta^2\sqrt{4\pi^2-\beta^2}$$