A cone shaped drinking cup is made from a circular piece of paper of radius r by cutting out a sector and joining the edges CB and CD. if the cut is made so that the volume of the resulting cup is maximised, then what is the ratio between the radius and height of the cup?

  • $\begingroup$ Please show what you've tried and where you're stuck to get better help from the community. $\endgroup$ – Saad Oct 9 '18 at 11:21
  • $\begingroup$ Interesting problem. Start with the equation for the volume of a cone and substitute the base radius for (xr) where x is the remaining fraction of the circle and write h in terms of (xr) and r? $\endgroup$ – Phil H Oct 9 '18 at 14:01

$V = \frac{1}{3}\pi R^2\cdot h$

$xC$ is the remaining circumference after removing a sector.

The base radius then becomes $\frac{xC}{2\pi} = \frac{x\cdot 2\pi\cdot r}{2\pi} = xr$

So $V = \frac{1}{3}\pi(rx)^2\cdot \sqrt{r^2-(rx)^2}$

$$\frac{dV}{dx} = \frac{2}{3}\pi\cdot r^2 x\cdot \sqrt{r^2-(rx)^2}+\frac{1}{3}\pi(rx)^2\cdot \frac{-r^2x}{\sqrt{r^2-(rx)^2}}$$

Max volume occurs when $\frac{dV}{dx} = 0$

Then $$\frac{2}{3}\pi\cdot r^2 x\cdot \sqrt{r^2-(rx)^2} = \frac{1}{3}\pi(rx)^2\cdot \frac{-r^2x}{\sqrt{r^2-(rx)^2}}$$

Which reduces to $2r^2 = 3(rx)^2$

And $x = \sqrt{\frac{2}{3}}$

Therefore the ratio of radius to height of the cone is $$\frac{rx}{h} = \frac{\sqrt{\frac{2}{3}}r}{\sqrt{r^2-(\sqrt{\frac{2}{3}}r)^2}} = \sqrt 2$$



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