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

Question

The following is the question :

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

The cylinder has a closed top and bottom.

2 formula :

(1) $V=r^2\pi h$

(2) $A=2r\pi h+2r^2\pi$ -> $A=2r\pi \left(h+r\right)$

I cannot find the equation for differentiation

How to find $A'$? Hints?

Thank your for your attention

From the volume formula, you have $ h = \dfrac {V}{\pi r^2} $. What can you do with this? — Ahaan S. Rungta, Nov 28 '13 at 16:35
Now I get $A=2r\pi \left(\frac{V}{r^2\pi }+r\right)$ but there are 2 variables (r & V)...so dA/d? Can I see V as a constant here? — Casper, Nov 28 '13 at 16:40

score 2 · Accepted Answer · answered Nov 28 '13 at 16:48

2

The volume is given, so that is a constant.

The volume constraint gives $h=\frac{V}{\pi r^2}$, from which we get $A(r) = 2 \pi r (r+\frac{V}{\pi r^2})$.

We see that $\lim_{r \downarrow 0} A(r) = \infty$ and $\lim_{r \to \infty} A(r) = \infty$, hence $A$ has a minimum.

Differentiate $A$ and set the derivative to zero. Solve for $r_0$. Then compute the corresponding $h_0$.

Details:

We have $A'(r) =-\frac{2\,\left( V-2\,\pi \,{r}^{3}\right) }{{r}^{2}}$, hence there is exactly one point $r_0$ for which $A'(r_0) = 0$, hence this must be he minimum. This gives$r_0 = \sqrt[3]{{V \over 2 \pi}}$. Then $h_0 = \frac{V}{\pi r_0^2} = \sqrt[3]{{4V \over \pi}}$.

answered Nov 28 '13 at 16:48

copper.hat

172,524

The answer in the book shows that h = 2r. Is it wrong? Anyway, thank you very much! – Casper Nov 28 '13 at 17:27
No, you can check that $h_0= 2 r_0$ above (remember that it is the 3rd root). – copper.hat Nov 28 '13 at 17:29
I mean the answer in the book seems wrong. I obtain the same answer as yours. – Casper Nov 28 '13 at 17:30
Both answers are correct. If you let $h=2r$ and solve for $r$ you will get the same formula. – copper.hat Nov 28 '13 at 17:54

score 0 · Answer 2 · answered Apr 17 '14 at 16:59

the question probably sounds like this: " A cylindrical container of radius r and height h has a constant volume V. The cost of materials for the surface of both its ends are twice the cost of its sides. State h in terms of r and V. Hence, find h and r in terms of V such that the cost is minimum."

Now, if you use the cost comparison given, it's 2(2*pi*r^2) = 2*pi*r*h which will conclude to 2r = h

since V = pi*r^2*h, state h in terms of V and r as V is constant, you'll get h = V/(pi*r^2)

Then, substitute 2r into h of h = V/(pi*r^2), now you get 2r = V/(pi*r^2), solve for the r and it will give r = (V/2*pi)^(1/3)

Now that you have r = (V/2*pi)^(1/3) , just substitute right hand side of r to the r in h = V/(pi*r^2) to get h which is h = (4V/pi)^(1/3)

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

2 Answers2