Find the radius of a cylinder of given volume V if its surface area is a minimum.

Question

this question is driving me crazy as I'm not sure how they've got the answer.

The surface area is given as $S = 2\pi r^2 + \frac {1}{50r} $ and they are asking for the value of r for which S is minimum.

The derivative of this (I hope!) is $4\pi r - \frac {1}{50r^2}$

Then to find the value of r when S is a minimum I presume you set the derivative to equal $0$.

The book shows the value $200π - \frac 13 $ but I'm not sure how they've got this figure from setting the derivative to $0$.

Any insight would be appreciated!

Edit: My bad, answer is raised to negative $ \frac {1}{3}$, you live, you learn. Thanks for pointing this out.

Answer 1

Let's remind ourselves that the the volume of a cylinder is $\pi r^2h$. Hence, if we set $$V=\pi r^2h$$then we can substitute $h$ for $r$ in the surface area formula using this relation.

Hence, Surface area of a cylinder is $2\pi r^2+2\pi rh$. After replacement, we get $$\textrm{Surface Area }=2\pi r^2 + \frac{2V}r$$We now take the derivative of this function, to get $$4\pi r -\frac{2V}{r^2}$$We set the derivative to zero, giving us the relation $2\pi r = \frac V{r^2}$ which means $$r^3=\frac V{2\pi}$$There is only one real solution for $r$ to this solution, which is $$r=\sqrt[3]{\frac V{2\pi}}$$We can check that this is the minimum by taking the second derivative of the Surface Area formula and plugging in this solution. $$4\pi+\frac{4V}{r^3}\to4\pi+8\pi=12\pi>0$$Since the second derivative is greater than zero, our $r$ value is indeed a minimum.

Now, let's try to plug in your values. Note that in your formula, instead of $\frac{2V}{r}$, you have $\frac1{50r}$. Hence, $$2V=\frac1{50}\to V=\frac1{100}$$Hence, our answer is $$\sqrt[3]{\frac1{200\pi}}=\frac1{10}\cdot\sqrt[3]{\frac1{2\pi}}$$

Answer 2

$$s=2\pi r^2+\dfrac{1}{50r}...........(1)$$ $$s'=4\pi r-\dfrac{1}{50r^2}$$ To be minimum, $s'=0$ So,$$4\pi r-\dfrac{1}{50r^2}=0$$ $$4\pi r=\dfrac{1}{50r^2}$$ $$\implies r^3=\dfrac{1}{50.4.\pi}$$ $$\implies r^3=\dfrac{1}{200\pi}$$ $$\implies r=\sqrt[3]{\dfrac{1}{200\pi}}$$ $$r=(200\pi)^{-\dfrac{1}{3}}$$ Now if we plus this value of r in the equation (1) and if the result is greater than zero,then it will be the minimum point. putting the value of r in equation (1),we get $$s=\dfrac{3}{10}\sqrt[3]{\dfrac{\pi}{5}}>0$$ So,in this point there is a minimum value of the function.

Answer 3

Use the inequality beetwen arithmetic and geometric mean:

$$S = 2\pi r^2 + \frac {2V}{r} =2\pi r^2+\frac {V}{r}+ \frac {V}{r}\geq 3\sqrt[3]{2\pi V^2}$$

and equality is achieved iff $r = \sqrt[3]{V\over 2\pi}$. So we have $\boxed{S_{\min} =3\sqrt[3]{2\pi V^2}}$

Answer 4

By the classical formulas, $$V=\pi r^2h$$ and $$S=2\pi r^2+2\pi rh.$$

Eliminating $h$,

$$S=2\pi r^2+\frac{2V}r.$$

Then cancelling the derivative,

$$4\pi r-\frac{2V}{r^2}=0$$ or $$r^3=\frac V{2\pi}.$$