Building a cone of maximum volume

Let $s$ be the radius of the circle, and $\phi$ the angle of the sector to be folded. Upon folding, the radius of the base $r$ is related to $s$ and $\phi$ by

$r = s \dfrac{\phi}{ 2 \pi}$

and therefore, the height of the cone will be $h = \sqrt{ s^2 - r^2 } = s \sqrt{ 1 - (\dfrac{\phi}{2 \pi})^2 }$

It follows that the volume of the cone is

$V = \dfrac{1}{3} \pi s^3 (\dfrac{\phi}{2 \pi} )^2 \sqrt{1 - (\dfrac{\phi}{2 \pi} )^2 }$

Let $x = \dfrac{\phi}{2 \pi}$ , then we want to maximize

$f(x) = x^2 \sqrt{1 - x^2}$

The derivative is given by

$f '(x) = 2 x (1 - x^2)^{1/2} - x^3 (1 - x^2)^{-1/2} = (1 - x^2)^{-1/2} (2 x - 3 x^3 )$

$f '(x) = 0$ implies $x = 0$ or $x = \sqrt{\dfrac{2}{3}}$

$x=0$ corresponds to zero volume and is therefore rejected. So the maximum occurs at $x = \sqrt{\dfrac{2}{3}}$

Which implies that $\phi = 2 \pi \sqrt{\dfrac{2}{3}} = 293.94^{\circ}$

It is easy to see, that the size of radius of the original circle shape (s, also the slant height of the cone) does not influence the size of the residual central angle of the cone with the maximal volume (this is due to similarity).

$\text {Let } s = 1$

(We could use this method as well when calculating lengths/areas/volumes with parameters or actual values, by using an enlargement, scale factor = s).

Then, we can take the (well known) right angled triangle formed by the radius of the base circle of the cone (r), the height of the cone (h) and the slant height of the cone (s = 1, hypotenuse) and apply Pythagoras' theorem:

$r^2 = 1- h^2$

The volume of the cone:

$V = \frac {\pi}{3} r^2 h = \frac {\pi}{3} (1 - h^2) h = \frac {\pi}{3} (h - h^3) \rightarrow \max$

The first derivative of V:

$\frac {\pi}{3} (1 - 3h^2) = 0$

$1 - 3h^2 = 0$

Since h is a height, h > 0:

$h = \sqrt { \frac {1}{3} }$

And it is easy to see from the second derivative of V, that this is really a maximum as:

$V'' = \frac {\pi}{3} (1 - 6h) = \frac {\pi}{3} (1 - 2 \sqrt {3} ) < 0$

Then:

$r = \sqrt { \frac {2}{3} }$

The residual central angle, x can be calculated by using the ratio between the circumferences of the original circle and the base circle of the cone (which is the same as the ratio of their radii, r and s (= 1)):

$x = \frac { r }{ s } × 360° = \sqrt { \frac {2}{3} } × 360° = \boxed { 293.94° \text { (2 d. p.) } }$

Since this is under the calculus section, i challenged myself to do this without looking up the volume of a cone, i did the following: Let the angle be theta, let the radius of the circle be a constant. The radius of the bottom section of the cone must then be

$R(\theta)=r\cdot\frac{\theta}{360}$

We can now imagine we have a triangle where the height of the cone is the base, the radius of the original circle is the hypothenuse, and the radius of the bottom of the cone is the last cathetus.

We can then find the height by using the pythagorean theorem

$h(\theta)=\sqrt{r^2-R^2(\theta)}$

If we now want a linear function that follows the edge of the cone, it must be

$f(x,\theta)=\frac{R(\theta)}{h(\theta)}*x$

The volume of the cone must then be a volume of revolution, given by

$V(\theta)=2\pi\cdot\int_{0}^{h(\theta)}f^2(x,\theta)dx$

We can now solve for

$V'(\theta)=0$

If we discard the 0, we find a maximum at 293.94