Maximum surface area of cylinder

[songoku]

Homework Statement

Given that a solid cylinder has a fixed volume V, prove that its total surface area S is minimum when its height and base diameter are equal.

Homework Equations

derivative

The Attempt at a Solution

I am able to prove that question.

$$V=\pi r^2 h$$

$$h=\frac{V}{\pi r^2}$$


So, to get minimum surface area:

$$\frac{dS}{dr}=0$$

$$\frac{d}{dr}(2\pi r h + 2 \pi r^2)=0$$

$$\frac{d}{dr}(2\pi r \frac{V}{\pi r^2} + 2 \pi r^2)=0$$

$$\frac{d}{dr}(2\frac{V}{r}+2 \pi r^2)=0$$

$$-2\frac{V}{r^2}+4\pi r=0$$

$$2\frac{V}{r^2}=4\pi r$$

$$2\frac{\pi r^2 h}{r^2}=4\pi r$$

$$h=d\; \text{(Shown)}$$

So,with h = d, the minimum surface area is :

$$S=2\pi r (2r) + 2\pi r^2$$

$$S=6\pi r^2$$

What I want to ask is : how about if the question asks to find the maximum surface area?

I think to find the maximum value, we also set $$\frac{dS}{dr}=0$$. From my work, I don't see any ways to find the maximum value...

Thanks

 

[ideasrule]

There is no maximum surface area. If you examine the equation for S that you derived:

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

As r gets very big, S approaches infinity. Similarly, as r gets very small, S also approaches infinity.

To visualize this, if you squash the cylinder so that it has a tiny height, the bases can be expanded to keep the volume constant. Since the bases can be expanded to any size, the maximum surface area is infinite. If you squeeze the cylinder so that it has a tiny radius but large height, surface area would also increase, but it's a bit harder to see this intuitively.

 

[tiny-tim]

What I want to ask is : how about if the question asks to find the maximum surface area?

I think to find the maximum value, we also set $$\frac{dS}{dr}=0$$. From my work, I don't see any ways to find the maximum value...

Thanks

Hi songoku!

Your proof is fine, but technically you haven't actually proved that r = d gives a minimum volume …

you've only proved that the volume has exactly one stationary value.

Now you must
i] show it's a minimum
ii] (to find the maxima) use the physical fact that d and h must both be greater than 0.

 

[songoku]

Hi ideasrule and tiny-tim :)

You're right tiny-tim. Technically I haven't actually proved that h = d gives a minimum volume. I tried but not sure if it's right. I used second derivative test.

$$\frac{dS}{dr}=-2\frac{V}{r^2}+4\pi r$$

$$\frac{d^2S}{dr^2}=4\frac{V}{r^3}+4\pi$$

$$\frac{d^2S}{dr^2}=4\frac{\pi r^2h}{r^3}+4\pi$$

$$\frac{d^2S}{dr^2}=\frac{4\pi h}{r}+4\pi$$For h = d :

$$\frac{d^2S}{dr^2}=12 \pi$$

Since $$\frac{d^2S}{dr^2}>0$$ , it's minimum value.For maxima, I think ideasrule is right. But I don't know how to use the physical fact that d and h must both be greater than 0 to get maxima value. Is it the same as ideasrule's work?

Thanks :)

 

[tiny-tim]

I used second derivative test.

$$\frac{dS}{dr}=-2\frac{V}{r^2}+4\pi r$$

$$\frac{d^2S}{dr^2}=4\frac{V}{r^3}+4\pi$$ …

eugh!

you should have stopped there! …

you don't need to know what d²S/dr² is, you only need to know whether it's positive! …

and it always is, since V and r are always positive!

For maxima, I think ideasrule is right. But I don't know how to use the physical fact that d and h must both be greater than 0 to get maxima value. Is it the same as ideasrule's work?

Since d²S/dr² > 0, it has only one minimum and no other turning-point for r > 0 …

so in the range r > 0, S has (or, rather, approaches) a maximum at the two end-points, 0 and ∞.

 

[songoku]

Hi tiny-tim

Oh I get it. Great thanks for your help !

And also thank you idearules.