# Thread: Function of a cylinder

1. ## Function of a cylinder

A can producer must make a cylindrical can that contains a volume of 16 ¶ cubic centimeters. Find the function this producer must analyze to make sure the amount of material is the least possible.

2. Originally Posted by bret80
A can producer must make a cylindrical can that contains a volume of 16 ¶ cubic centimeters. Find the function this producer must analyze to make sure the amount of material is the least possible.
$V=\frac{1}{3}\pi r^2 h$
The amount of matherial used is,
$A=2\pi r h+2\pi r^2=$
You want to minimize $A$.

Note in first equation we have,
$\frac{3V}{h\pi}=r^2$
Thus,
$A=2\pi \sqrt{\frac{3V}{h\pi }} h+2\pi \frac{3V}{h\pi}$
Simplify (steps omitted),
$2\sqrt{3V\pi}\cdot \sqrt{h}+\frac{6V}{h}$
Derivative time,
$A'=\sqrt{3V\pi}h^{-1/2}-6Vh^{-2}$
Make equal to zero,
$\sqrt{3V\pi}h^{-1/2}=6Vh^{-2}$
Divide both sides by $h^{-2}$,
$\sqrt{3V\pi}h^{3/2}=6V$
Thus,
$h^{3/2}=\frac{1}{\sqrt{\pi}}2\sqrt{3}\sqrt{V}$
Note we substitute the value for "V":
$h^{3/2}=\frac{2\sqrt{3}\cdot 4}{\sqrt{\pi}}$
Thus,
$h^{3/2}=\frac{8\sqrt{3}}{\sqrt{\pi}}$
Square both sides,
$h^3=\frac{2^6 \cdot 3}{\pi}$
Now take cube root,
$h=\frac{4 \cdot \sqrt[3]{3}}{\sqrt[3]{\pi}}$

3. The correct answer I have is

32/r + 2¶r^2

4. Originally Posted by bret80
The correct answer I have is

32/r + 2¶r^2
\

This time, TPH, you've gone too far!!

(All he wanted was the equation to minimize. )

bret80: Follow the first two lines of TPH's thread, but instead of solving the Volume equation for r, solve it for h. Sub that value for h into the Area equation.

-Dan

5. Originally Posted by ThePerfectHacker
$V=\frac{1}{3}\pi r^2 h$
Isn't that the volume of a cone?

RonL

6. Hello, Bret!

The correct answer I have is: 32/r + 2πr² . . . . not quite

The volume of a cylinder is: . $V\:=\:\pi r^2h$
We are told that the volume is $16\pi$ cm³.

So we have: . $\pi r^2h \:=\:16\pi\qiad\Rightarrow\quad h \:=\:\frac{16}{r^2}$ [1]

The surface area of a cylinder is: .top + bottom + side.

The top and bottom are circles of radius $r$. .Their area is: $2 \times \pi r^2$

The side is a rectangle of length $2\pi r$ and height $h$: area = $2\pi rh$

Hence, the surface area of a cylinder is: . $S \;=\;2\pi r^2 + 2\pi rh$

Substitute [1]: . $S\;=\;2\pi r^2 + 2\pi r\left(\frac{16}{r^2}\right)$

Therefore: . $\boxed{S \;= \;2\pi r^2 + \frac{32\pi}{r}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

By the way, you can use &# 960; (without the space) to make π.

More codes (until you learn LaTeX):

&# 178; . . squared: ²

&# 179; . . cubed: ³

&# 8776; . approx: ≈

&# 8800; . not equal: ≠

&# 952; . . theta: θ

&# 177; . . plus/minus: ±

&# 176; . . degree: °

&# 8730; . radical: √

&# 183; . . dot: ·

&# 189; . . one-half: ½