# Thread: Minimizing surface area...one step further

1. ## Minimizing surface area...one step further

This is taking the classic 'minimizing the surface area of a cylinder' problem one step further. In the first part of the problem, it was given that the volume is 1000mL. I found the height and radius that would minimize the surface area.

The second part of the problem is this:

The material for the cans is cut from sheets of metal. The cylindrical sides are
formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if
the top and bottom discs are cut from squares of side 2r, this leaves considerable waste metal, which
may be recycled but has little or no value to the can makers. If this is the case, show that the total
amount of metal used (including the waste metal) is minimized when h/r is 8/(pi)

The task is to show that the ratio of height to radius is 8pi).

How is this problem done?

2. The amount of metal used is $2\pi r h + 2 \times (2r)^2$. If the volume is fixed at C, then $h = C/ \pi r^2$ and $h/r = C/ \pi r^3$. So we have the metal used is $2 C / r + 8r^2$. Differentiating, we find that the minimum occurs at $r^3 = C/8$ (check that this is a minimum). Substituting, $h/r = 8/\pi$.

3. Originally Posted by bandito1989
This is taking the classic 'minimizing the surface area of a cylinder' problem one step further. In the first part of the problem, it was given that the volume is 1000mL. I found the height and radius that would minimize the surface area.

The second part of the problem is this:

The material for the cans is cut from sheets of metal. The cylindrical sides are
formed by bending rectangles; these rectangles are cut from the sheet with little or no waste. But if
the top and bottom discs are cut from squares of side 2r, this leaves considerable waste metal, which
may be recycled but has little or no value to the can makers. If this is the case, show that the total
amount of metal used (including the waste metal) is minimized when h/r is 8/(pi)

The task is to show that the ratio of height to radius is 8pi).

How is this problem done?
Hi bandito1989,

you can start by writing h in terms of r as the volume is 1000 cubic units.

${\pi}r^2h=1000$

$h=\frac{1000}{{\pi}r^2}$

The metal used is $2{\pi}rh+8r^2=\frac{2{\pi}r(1000)}{{\pi}r^2}+8r^2= \frac{2000}{r}+8r^2=\frac{2000+8r^3}{r}$

Finding the r giving minimum

$\frac{d}{dr}\ \frac{2000+8r^3}{r}=\frac{r(24r^2)-\left(2000+8r^3\right)}{r^2}$

$=\frac{24r^3-8r^3-2000}{r^2}=\frac{16r^3-2000}{r^2}$

This is zero when the numerator is zero

$16r^3=2000\ \Rightarrow\ r^3=\frac{2000}{16}=\frac{500}{4}=125$

$r=5$

$h=\frac{1000}{{\pi}r^2}=\frac{1000}{25{\pi}}=\frac {40}{{\pi}}$

$\frac{h}{r}=\frac{40}{5{\pi}}=\frac{8}{{\pi}}$

4. Hi! Thank you so much, Archie Meade!
I'm trying to do part two of the problem, but I'm having trouble figuring it out. I think I've set up the problem correctly, but it doesn't work. Could you help me?

Problem: A more efficient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons. Show that if this strategy is adopted, then h/r = 4rt3 / pi ~ 2.21.

I have V = 1000 = pir^2h
and
A = 2pirh + 2(.5ap) with a being the apothem and p being the perimeter. I found the perimeter to be 12* a * tan(30). Am I on the right track? Or... far from it? Any help would be great! Thank you!

5. Originally Posted by spirits003
Hi! Thank you so much, Archie Meade!
I'm trying to do part two of the problem, but I'm having trouble figuring it out. I think I've set up the problem correctly, but it doesn't work. Could you help me?

Problem: A more efficient packing of the discs is obtained by dividing the metal sheet into hexagons and cutting the circular lids and bases from the hexagons. Show that if this strategy is adopted, then h/r = 4rt3 / pi ~ 2.21.

I have V = 1000 = pir^2h
and
A = 2pirh + 2(.5ap) with a being the apothem and p being the perimeter. I found the perimeter to be 12* a * tan(30). Am I on the right track? Or... far from it? Any help would be great! Thank you!
hi spirits003,

yes, you're on the right track...

remember that the apothem is the radius of the circle,
if the circles fit exactly within the hexagons against the hexagon sides.

$Total\ surface\ area=2{\pi}rh+(2)6a^2tan30^o=2{\pi}rh+12r^2\ \frac{1}{\sqrt{3}}$

$=2{\pi}rh+\frac{12r^2}{\sqrt{3}}=\frac{2000}{r}+\f rac{12r^2}{\sqrt{3}}$

$=\frac{\sqrt{3}(2000)+12r^3}{\sqrt{3}r}$

Differentiating this and equating to zero

$\frac{d}{dr}\left(\frac{u}{v}\right)=\frac{\sqrt{3 }r\left(36r^2\right)-\left(\sqrt{3}(2000)+12r^3\right)\sqrt{3}}{3r^2}$

For this to be zero, the numerator must be zero

$36\sqrt{3}r^3=6000+\sqrt{3}(12)r^3\ \Rightarrow\ 6\sqrt{3}r^3=1000+2\sqrt{3}r^3$

$4\sqrt{3}r^3=1000$

$r^3=\frac{1000}{4\sqrt{3}}$

$h=\frac{1000}{{\pi}r^2}\ \Rightarrow\ \frac{h}{r}=\frac{1000}{{\pi}r^3}=\frac{1000}{{\pi }}\ \frac{1}{r^3}=\frac{1000}{{\pi}}\ \frac{4\sqrt{3}}{1000}$

$=\frac{4\sqrt{3}}{{\pi}}$

6. ## Archie Meade is THE best :)

Thank you SO much! You right SUCH A LIFE-SAVER!
I actually went further than what I wrote earlier, and I knew that the apothem was the radius. I apparently can't do simple arthimetic / differeniate ><

I redid the problem myself just now (using your work as confirmation), and I ACTUALLY got the right answer this time. I've been struggling (for no good reason either) with this problem for days, literally. And now, it's SO crystal clear! Thank you SO, SO, SO much!!! This is SUCH a relief!

I wish I could bring you to everywhere I am when I have to do Math! Haha.

Could I ask you a question if you don't mind? Are you a student or professor/teacher or just a person who really loves math? I'm just really curious. All of your steps are so clear and wonderful.

7. Originally Posted by spirits003

Thank you SO much! You right SUCH A LIFE-SAVER!
I actually went further than what I wrote earlier, and I knew that the apothem was the radius. I apparently can't do simple arthimetic / differeniate ><

I redid the problem myself just now (using your work as confirmation), and I ACTUALLY got the right answer this time. I've been struggling (for no good reason either) with this problem for days, literally. And now, it's SO crystal clear! Thank you SO, SO, SO much!!! This is SUCH a relief!

I wish I could bring you to everywhere I am when I have to do Math! Haha.

Could I ask you a question if you don't mind? Are you a student or professor/teacher or just a person who really loves math? I'm just really curious. All of your steps are so clear and wonderful.
Hi spirits003,

I do work as a freelance mathematics instructor.
I offer services to students who want to score particular grades
or who want to learn maths enough to get through their exams.

I write my own teaching materials for them as the school textbooks here
are not very instructional and half the teachers are unqualified to teach basic math anyway.

Yes, i like maths very much,
but it pleases me to assist pre-university students to get high grades that they are happy with.

I also train electronics graduates and design tattoos!!