# How to minimize the cost of a silo

## Homework Equations

Maxima/minima are where the first derivative is 0
Volue of a hemisphere: ##~\displaystyle V=\frac{2}{3}\pi r^3##
Area of a hemisphere: ##~\displaystyle A=2\pi r^2##

## The Attempt at a Solution

$$A=2\pi rh+2\pi r^2=2\pi[h+r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{2r}{3}-\frac{V}{\pi r^2}$$
$$A=2\pi r\left[ \frac{2r}{3}-\frac{V}{\pi r^2}+r \right]=2\pi r\left[ \frac{5}{3}r-\frac{V}{\pi r^2} \right]$$
$$A'=2\pi \left[ \frac{5}{3}r-\frac{V}{\pi r^2}+r\left( \frac{5}{3}+\frac{2V}{\pi}r^{-3}\right)\right]$$
$$A'=0:~2\pi\left[ \frac{10}{3}r-\frac{V}{\pi r^2}+\frac{2V}{\pi r^3}\right]=0$$
$$2\pi\left[ \frac{10\pi r^3-3Vr+6V}{3\pi r^3} \right]=0$$
$$\Rightarrow~10\pi r^3=3V(r-2)$$
I express V in terms of ##~\xi=\frac{h}{r}~##:
$$V=\pi r^3\xi+\frac{2}{3}\pi r^3$$
$$10\pi r^3=3\left[ \pi r^3\xi+\frac{2}{3}\pi r^3 \right](r-2)$$
Blocked end

#### Attachments

• Capture.JPG
15.3 KB · Views: 318
• Capture2.JPG
3.8 KB · Views: 277
• Capture.JPG
24.3 KB · Views: 599
• Capture2.JPG
3.8 KB · Views: 593

Orodruin
Staff Emeritus
Homework Helper
Gold Member
First of all, you are not taking into account that the hemisphere costs more per surface area to construct than the cylinder. You are just computing the surface area.

Second, you have made an arithmetic error somewhere because your dimensions do not match.

epenguin
Homework Helper
Gold Member

## Homework Statement

View attachment 222141
View attachment 222142

## Homework Equations

Maxima/minima are where the first derivative is 0
Volue of a hemisphere: ##~\displaystyle V=\frac{2}{3}\pi r^3##
Area of a hemisphere: ##~\displaystyle A=2\pi r^2##

## The Attempt at a Solution

$$A=2\pi rh+2\pi r^2=2\pi[h+r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{2r}{3}-\frac{V}{\pi r^2}$$
$$A=2\pi r\left[ \frac{2r}{3}-\frac{V}{\pi r^2}+r \right]=2\pi r\left[ \frac{5}{3}r-\frac{V}{\pi r^2} \right]$$
$$A'=2\pi \left[ \frac{5}{3}r-\frac{V}{\pi r^2}+r\left( \frac{5}{3}+\frac{2V}{\pi}r^{-3}\right)\right]$$
$$A'=0:~2\pi\left[ \frac{10}{3}r-\frac{V}{\pi r^2}+\frac{2V}{\pi r^3}\right]=0$$
$$2\pi\left[ \frac{10\pi r^3-3Vr+6V}{3\pi r^3} \right]=0$$
$$\Rightarrow~10\pi r^3=3V(r-2)$$
I express V in terms of ##~\xi=\frac{h}{r}~##:
$$V=\pi r^3\xi+\frac{2}{3}\pi r^3$$
$$10\pi r^3=3\left[ \pi r^3\xi+\frac{2}{3}\pi r^3 \right](r-2)$$
Blocked end

There appears to be an error – omission of factor r - after your second = , maybe this is only transcription and does not carry through.

You are likely to get into confusion or mistakes by using the same symbols V and A each to represent two different things. I think you have already got incompatible equations in there. A mistake I think you have made before.

Then,I have not followed what you actually doing, but what you essentially say you are doing by A' is to minimise A the area (one of them) but the problem is to minimise the cost which is a different formula.

(May be some posting overlap here).

A good approach but there are two things to note :
1. You're not trying to minimize the area, you're minimizing cost. In a way, you're minimizing some "effective area" function.
2. Have you ever encountered Lagrange multipliers?

Also, as others mentioned, check the algebra, but if you do get a convoluted polynomial for r, there's no shame in using a computer to find its roots.

I denote the cost as C:
$$\text{[1]}~ C=2\pi rh+4\pi r^2=2\pi r[h+2r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{2r}{3}-\frac{V}{\pi r^2}$$
$$\text{[2]}~C=2\pi r\left[ \frac{2r}{3}-\frac{V}{\pi r^2}+2r \right]=2\pi r\left[ \frac{8}{3}r-\frac{V}{\pi r^2} \right]$$
$$\text{[3]}~C'=2\pi \left[ \frac{8}{3}r-\frac{V}{\pi r^2}+r\left( \frac{8}{3}+\frac{2V}{\pi}r^{-3}\right)\right]$$
$$\text{[4]}~C'=0:~2\pi\left[ \frac{16}{3}r-\frac{V}{\pi r^2}+\frac{2V}{\pi r^3}\right]=0$$
$$\text{[5]}~2\pi\left[ \frac{16\pi r^4-3Vr+6V}{3\pi r^3} \right]=0$$
$$\text{[6]}~\Rightarrow~10\pi r^4=3V(r-2)$$
I don't know the Lagrange multipliers and i think i needn't know them since i follow a basic algebra book from the beginning and they weren't there so far
The units in eq. [6] match but no solution
Indeed i used A meaning the cost function, but V was always for volume

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You still have dimensional errors, which means you are doing your arithmetic wrong. Please correct it.
The units in eq. [6] match
No they don’t.

You mean that i can't have (r-2) since 2 is dimensionless?

Ray Vickson
Homework Helper
Dearly Missed
I denote the cost as C:
$$\text{[1]}~ C=2\pi rh+4\pi r^2=2\pi r[h+2r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{2r}{3}-\frac{V}{\pi r^2}$$
$$\text{[2]}~C=2\pi r\left[ \frac{2r}{3}-\frac{V}{\pi r^2}+2r \right]=2\pi r\left[ \frac{8}{3}r-\frac{V}{\pi r^2} \right]$$
$$\text{[3]}~C'=2\pi \left[ \frac{8}{3}r-\frac{V}{\pi r^2}+r\left( \frac{8}{3}+\frac{2V}{\pi}r^{-3}\right)\right]$$
$$\text{[4]}~C'=0:~2\pi\left[ \frac{16}{3}r-\frac{V}{\pi r^2}+\frac{2V}{\pi r^3}\right]=0$$
$$\text{[5]}~2\pi\left[ \frac{16\pi r^4-3Vr+6V}{3\pi r^3} \right]=0$$
$$\text{[6]}~\Rightarrow~10\pi r^4=3V(r-2)$$
I don't know the Lagrange multipliers and i think i needn't know them since i follow a basic algebra book from the beginning and they weren't there so far
The units in eq. [6] match but no solution
Indeed i used A meaning the cost function, but V was always for volume

In the first line your expression for ##h## in terms of ##r## and ##V## is incorrect; ##V## should have a positive coefficient and ##r## a negative coefficient.

$$\text{[1]}~ C=2\pi rh+4\pi r^2=2\pi r[h+2r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{V}{\pi r^2}-2r$$
$$\text{[2]}~C=2\pi r\left[ \frac{V}{2\pi r^2}-2r+2r \right]=\frac{2V}{r}$$
$$C'=-\frac{2V}{r^2}$$
No

Ray Vickson
Homework Helper
Dearly Missed
$$\text{[1]}~ C=2\pi rh+4\pi r^2=2\pi r[h+2r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{V}{\pi r^2}-2r$$
$$\text{[2]}~C=2\pi r\left[ \frac{V}{2\pi r^2}-2r+2r \right]=\frac{2V}{r}$$
$$C'=-\frac{2V}{r^2}$$
No

You still persist in making elementary algebra errors that render all you later work invalid. Start again, and BE CAREFUL.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
You mean that i can't have (r-2) since 2 is dimensionless?
Yes. And as @Ray Vickson says, all of your errors - in this thread and in others - could have been avoided if you were simply more careful when doing your arithmetic.

$$C=2\pi rh+4\pi r^2=2\pi r[h+2r],~~V=\pi r^2h+\frac{2}{3}\pi r^3~\rightarrow~h=\frac{V}{\pi r^2}-\frac{2}{3}r$$
$$C=2\pi r\left[\frac{V}{\pi r^2}-\frac{2}{3}r+2r\right]=2\pi r\left[\frac{V}{\pi r^2}+\frac{4}{3}r\right]$$
$$C'=2\pi \left[\frac{V}{\pi r^2}+\frac{4}{3}r+r\left( (-2)\frac{V}{2\pi r^2}r^{-3}+\frac{4}{3} \right) \right]=2\pi\left[\frac{8}{3}r-\frac{V}{\pi r^2}\right]$$
$$C'=0~\rightarrow~r^3=\frac{3V}{8\pi}$$

I have the number 8 in the denominator that's redundant

#### Attachments

• Capture.JPG
1.8 KB · Views: 178
• Capture2.JPG
3.2 KB · Views: 167
Orodruin
Staff Emeritus
Homework Helper
Gold Member
You are aware that the radius is half the diameter, right?

A bit of humor won't hurt...
About the h, i don't see it equals the diameter:
$$r=\frac{1}{2}\sqrt[3]{\frac{3V}{\pi}}$$
$$h=\frac{V}{\pi r^2}-\frac{2}{3}r=\frac{V}{\frac{\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}-\frac{2}{3}\frac{1}{2}\sqrt[3]{\frac{3V}{\pi}}$$
This won't cancel to become 2r=h

You are aware that the radius is half the diameter, right?
Sorry, relatively new to the forums. Is your attitude the usual around here?

Orodruin, i liked your subtle humor, i laughed.
I hope you didn't offend, since i see others might:
Sorry, relatively new to the forums. Is your attitude the usual around here?

$$h=\frac{V}{\pi r^2}-\frac{2}{3}r=\frac{V}{\frac{\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}-\frac{2}{3}\frac{1}{2}\sqrt[3]{\frac{3V}{\pi}}$$
$$h=\frac{V-\frac{\pi}{4}\sqrt[3]{\frac{3V}{\pi}}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}{3\frac{\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}=$$
$$h=\frac{V-\frac{\pi}{4}\frac{3V}{\pi}}{\frac{3\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}=\frac{\frac{1}{4}V}{\pi\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}$$
$$=\frac{V^{(1-2/3)}}{4\pi\sqrt[3]{\left( \frac{3}{\pi} \right)^2}}=\frac{\sqrt[3]{V}}{4\pi\sqrt[3]{\left( \frac{3}{\pi} \right)^2}}$$
It doesn't equal r, so h≠r and it should

After a few trials i got the right answer, that the height of the cylinder equals the diameter:
$$h=\frac{V}{\pi r^2}-\frac{2}{3}r=\frac{V}{\frac{\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}-\frac{2}{3}\frac{1}{2}\sqrt[3]{\frac{3V}{\pi}}$$
$$h=\frac{3V-\frac{\pi}{4}\sqrt[3]{\frac{3V}{\pi}}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}{3\frac{\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}=\frac{3V-\frac{\pi}{4}\frac{3V}{\pi}}{\frac{3\pi}{4}\sqrt[3]{\left( \frac{3V}{\pi} \right)^2}}$$
$$=\frac{\frac{9}{4}V}{\frac{3\pi\sqrt[3]{(3V)^2}}{4\sqrt[3]{\pi^2}}}=\frac{3V}{\sqrt[3]{\pi}\sqrt[3]{(3V)^2}}=\sqrt[3]{\frac{3V}{\pi}}=2r$$

Orodruin
PeterDonis
Mentor
2020 Award
Moderator's note: Thread title changed to be more descriptive and less generic.