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?
[Math] Find ratio between radius and height of a cone with maximised volume
calculus
Related Solutions
Let $r$ be the radius & $h$ be the height of the cylinder having its total surface area $A$ (constant) since cylindrical container is closed at the top (circular) then its surface area (constant\fixed) is given as $$=\text{(area of lateral surface)}+2\text{(area of circular top/bottom)}$$$$A=2\pi rh+2\pi r^2$$ $$h=\frac{A-2\pi r^2}{2\pi r}=\frac{A}{2\pi r}-r\tag 1$$
Now, the volume of the cylinder $$V=\pi r^2h=\pi r^2\left(\frac{A}{2\pi r}-r\right)=\frac{A}{2}r-\pi r^3$$ differentiating $V$ w.r.t. $r$, we get $$\frac{dV}{dr}=\frac{A}{2}-3\pi r^2$$ $$\frac{d^2V}{dr^2}=-6\pi r<0\ \ (\forall\ \ r>0)$$ Hence, the volume is maximum, now, setting $\frac{dV}{dr}=0$ for maxima $$\frac{A}{2}-3\pi r^2=0\implies \color{red}{r}=\color{red}{\sqrt{\frac{A}{6\pi}}}$$ Setting value of $r$ in (1), we get $$\color{red}{h}=\frac{A}{2\pi\sqrt{\frac{A}{6\pi}}}-\sqrt{\frac{A}{6\pi}}=\left(\sqrt{\frac{3}{2}}-\frac{1}{\sqrt 6}\right)\sqrt{\frac{A}{\pi}}=\color{red}{\sqrt{\frac{2A}{3\pi}}}$$ Hence, the ratio of height $(h)$ to the radius $(r)$ is given as $$\color{}{\frac{h}{r}}=\frac{\sqrt{\frac{2A}{3\pi}}}{\sqrt{\frac{A}{6\pi}}}=\sqrt{\frac{12\pi A}{3\pi A}}=2$$ $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\frac{h}{r}=2}}$$
Below is a (not-to-scale and I-totally-didn't-misread-the-question-at-first-as-6-being-the-diameter-instead-of-the-radius) picture of the cone and cylinder in question (behold my abysmal drawing skills in Microsoft Paint thanks to my expired license on Mathematica!)
In any case, we can let the radius of the inscribed cylinder be $x$. That must therefore mean that the height of the cylinder is $4 - \frac{4}{6}x$. This makes the volume of the cylinder
$$V = \pi x^2 \left(4 - \frac{2}{3} x\right) = \pi \left(4x^2 - \frac{2}{3} x^3\right)$$
To find the maximum volume of the cylinder we take the derivative of $V$ with respect to $x$ and set it to $0$:
$$\frac{dV}{dx} = \pi \left(8x - 2x^2\right) = 0 \Rightarrow 8x - 2x^2 = 2x(4-x) = 0$$
$x = 0$ leads to a trivial case, which leaves us with $x = 4$. This means that the height is $4 - \frac{2}{3} \cdot 4 = \frac{4}{3}$, and the volume is therefore $$V = \pi \cdot 4^2 \cdot \frac{4}{3} = \frac{64}{3}\pi$$
Best Answer
$V = \frac{1}{3}\pi R^2\cdot h$
$xC$ 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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