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.
$\displaystyle V=\frac{1}{3}\pi r^2 h$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,
$\displaystyle h=\frac{4 \cdot \sqrt[3]{3}}{\sqrt[3]{\pi}}$
\Originally Posted by bret80
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
Hello, Bret!
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: .$\displaystyle \boxed{S \;= \;2\pi r^2 + \frac{32\pi}{r}}$
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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