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.

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- Aug 2nd 2006, 01:58 PMbret80Function 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.

- Aug 2nd 2006, 02:20 PMThePerfectHackerQuote:

Originally Posted by**bret80**

The amount of matherial used is,

$\displaystyle A=2\pi r h+2\pi r^2=$

You want to minimize $\displaystyle A$.

Note in first equation we have,

$\displaystyle \frac{3V}{h\pi}=r^2$

Thus,

$\displaystyle A=2\pi \sqrt{\frac{3V}{h\pi }} h+2\pi \frac{3V}{h\pi}$

Simplify (steps omitted),

$\displaystyle 2\sqrt{3V\pi}\cdot \sqrt{h}+\frac{6V}{h}$

Derivative time,

$\displaystyle A'=\sqrt{3V\pi}h^{-1/2}-6Vh^{-2}$

Make equal to zero,

$\displaystyle \sqrt{3V\pi}h^{-1/2}=6Vh^{-2}$

Divide both sides by $\displaystyle h^{-2}$,

$\displaystyle \sqrt{3V\pi}h^{3/2}=6V$

Thus,

$\displaystyle h^{3/2}=\frac{1}{\sqrt{\pi}}2\sqrt{3}\sqrt{V}$

Note we substitute the value for "V":

$\displaystyle h^{3/2}=\frac{2\sqrt{3}\cdot 4}{\sqrt{\pi}}$

Thus,

$\displaystyle h^{3/2}=\frac{8\sqrt{3}}{\sqrt{\pi}}$

Square both sides,

$\displaystyle h^3=\frac{2^6 \cdot 3}{\pi}$

Now take cube root,

$\displaystyle h=\frac{4 \cdot \sqrt[3]{3}}{\sqrt[3]{\pi}}$ - Aug 2nd 2006, 02:49 PMbret80
The correct answer I have is

32/r + 2¶r^2 - Aug 2nd 2006, 03:03 PMtopsquarkQuote:

Originally Posted by**bret80**

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

(All he wanted was the equation to minimize. :D )

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 - Aug 2nd 2006, 08:20 PMCaptainBlackQuote:

Originally Posted by**ThePerfectHacker**

RonL - Aug 3rd 2006, 04:10 AMSoroban
Hello, Bret!

Quote:

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

The volume of a cylinder is: .$\displaystyle V\:=\:\pi r^2h$

We are told that the volume is $\displaystyle 16\pi$ cm³.

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

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

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

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

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

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

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

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

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