# Thread: Optimization of a cylinder

1. ## Optimization of a cylinder

A craftswoman wants to make a cylindrical jewelry box that has volume, V, equal to 55 cubic inches.

She will make the base and side of the box out of a metal that costs 10 cents per square inch. The lid of the box will be made from a metal with a more ornate finish which costs 100 cents per square inch.
Writing the radius of the cylindrical box as r, and the height of the box as h, calculate the cost, C, in cents, of the metal used to produce the box in terms of h and r.

So I thought it would be something like

$10+100(2pir^2) + 10(2pirh)$
$110(2pir^2)+10(2pirh)$

but that doesn't seem to be right. (I did also try it with decimals, as it's in cents)

I'm sure I'm missing something obvious here, maybe someone can help???

EDIT: I figured it out, it was something obvious :I
But now I have a new question!

How do I go about differentiating this?

$100pir^2+10pir^2+10(2pi(55/(pir^2)))$

I got as far as

$200pir+20pir$

and thought that the rest would just be +10, but that's not it.

2. ## Re: Optimization of a cylinder

let h = 55/pi(r^2)
surface area without expensive lid = pir^2 + 2pirh, substitute in value for h
get the derivative of this and let it equal to zero to find the minumum value of r

The rest should be a cake walk

3. ## Re: Optimization of a cylinder

Originally Posted by Elemiah
A craftswoman wants to make a cylindrical jewelry box that has volume, V, equal to 55 cubic inches.

She will make the base and side of the box out of a metal that costs 10 cents per square inch. The lid of the box will be made from a metal with a more ornate finish which costs 100 cents per square inch.
Writing the radius of the cylindrical box as r, and the height of the box as h, calculate the cost, C, in cents, of the metal used to produce the box in terms of h and r.

So I thought it would be something like

$10+100(2pir^2) + 10(2pirh)$
Where did that leading "10" come from? Also you seem to be misunderstanding the statement of the problem. The top, but not the bottom, is made from the more expensive metal. Taking r to be the radius of the cylinder, the top has area $\pi r^2$ and so its cost will be $100\pi r^2$. The base also has area $\pi r^2$ so the cost of the bottom wil be $10\pi r^2$. Yes, the curved area of the cylinder is $2\pi rh$ so the cost of that will be $20\pi rh$. The total cost will be $100\pi r^2+ 10\pi r^2+ 20\pi rh= 110\pi r^2+ 20\pi rh$. As
MathJack said, since the volume is $\pi r^2h= 55$, you can replace h by $\frac{55}{\pi r^2}$ so the cost function, in terms of r only, is $110\pi r^2+ \frac{1100}{r}$.

Differentiate that, with respect to r, and set the derivative equal to 0.

$110(2pir^2)+10(2pirh)$

but that doesn't seem to be right. (I did also try it with decimals, as it's in cents)

I'm sure I'm missing something obvious here, maybe someone can help???

EDIT: I figured it out, it was something obvious :I
But now I have a new question!

How do I go about differentiating this?

$100pir^2+10pir^2+10(2pi(55/(pir^2)))$
Do you know how to differentiate $r^2$? What about $\frac{1}{r^2}= r^{-2}$?

I got as far as

$200pir+20pir$

and thought that the rest would just be +10, but that's not it.