Maximizing the total volume of multiple cones cut from a disk. I was reviewing for the A.P. calculus exam, when I got to maximization/minimization problems. I cam across this problem http://jwilson.coe.uga.edu/EMT668/EMAT6680.Folders/Howard/MaximumVol.Cone/MaxVol.CONE.slh.html (not here, in my calculus book), and solved it. I wondered what the solution would be if you could cut it into as many sectors as you want, and you turned all of them into cones. However, I was not sure how to solve it. 
Here is the problem stated. You are given a paper disk. You can cut multiple lines from the edge of the disk to the center to leave you with multiple sectors. You then roll up each sector into a cone. How many sector should you cut the disk into, and how big should each sector be to maximize the total volume of all the cones? 
 A: The volume of a cone obtained from a sector with radius $r$ and normalised central angle $x=\alpha/(2\pi)$ is given by:
$$
V(x)=x^2\sqrt{1-x^2},
$$
where $0\le x\le1$ and I discarded an inessential multiplicative constant of $\pi r^3/3$.
If we have more than two cones, we can ask whether unfolding two cones, and making from the sum of their sectors a single cone, leads to a cone with a volume larger than the sum of the original volumes. That happens for sector angles $x$ and $y$ such that $V(x+y)>V(x)+V(y)$, that is when:
$$
F(x,y)={V(x+y)\over V(x)+V(y)} >1.
$$
I plotted (with Mathematica) a contour-plot of $F(x,y)$:

As you can see, $F(x,y)\ge1$ if $x+y\le0.8$. But with three or more cones, the sum of the two smaller normalised central angles cannot be more than $2/3$. Hence merging the two smaller cones into a single cone will always lead to a greater total volume, and we then obtain the maximum volume with just two cones.
The solution for two cones is well known: the sum of the volumes is given by 
$f(x)=V(x)+V(1-x)$ whose derivative is
$$
f'(x)=2 x \sqrt{1-x^2}-\frac{x^3}{\sqrt{1-x^2}}
+\frac{(1-x)^3}{\sqrt{x(2-x)}}-2 (1-x)\sqrt{x(2-x)} .
$$
Equating that to zero, and eliminating square roots, leads to a cubic equation for $(x-1/2)^2$ with three real solutions, in addition to the "trivial" solution $x=1/2$ (which corresponds to a local minimum). The maximum occurs for $x\approx 0.324014$ and its symmetric $1-x$, with $f(x)\approx 0.43606$.
A: Let $r$ be the radius of the circle. Suppose that $\theta\in [0,360]$ is the angle of a sector that we cut. Let $\alpha:=\theta/360\in [0,1]$. Then, the circular base of the associated cone constructed by this sector has circumference $2\pi r \alpha,$ so that its radius is $\bar r=r\alpha$. The height of this cone is $h=\sqrt{r^2-\bar r^2}=\sqrt{r^2-r^2\alpha^2}=r\sqrt{1-\alpha^2}$. Thus, its volume is $$V(\alpha)=\frac{\pi}{3} \bar r^2h=
\frac{\pi}{3} r^2\alpha^2r\sqrt{1-\alpha^2}=
\frac{\pi }{3}r^3 \alpha^2\sqrt{1-\alpha^2}.$$
Thus, the problem of interest becomes as follows:
$$\max_{n\in \mathbb N\setminus \{1\}} \max_{\alpha_1,\alpha_2,\dots,\alpha_n} \sum_{i=1}^n V(\alpha_i) \quad \text{subject to} \quad \sum_{i=1}^n \alpha_i=1 \quad \text{ and } \quad \alpha_i\ge 0, \, \ \forall i\in [n],$$
which, by discarding constants, reduces to 
$$\max_{n\in \mathbb N\setminus \{1\}} \max_{\alpha_1,\alpha_2,\dots,\alpha_n} \sum_{i=1}^n \alpha_i^2\sqrt{1-\alpha_i^2}. \quad \text{subject to} \quad \sum_{i=1}^n \alpha_i=1 \quad \text{ and } \quad \alpha_i\ge 0, \, \ \forall i\in [n].$$
So, the main job is to solve the interior problem!
