# I need to minimize the surface area of this cylinder

• November 23rd 2008, 02:15 PM
la_lo626
I need to minimize the surface area of this cylinder
design a cylinder which can hold a volume of 10ft cubed. so that it's used with minimal material. what are the dimensions (r and h) and what is the relationship between r and h. HINT: MINIMIZE SURFACE AREA????????????
• November 23rd 2008, 03:15 PM
TheEmptySet
Quote:

Originally Posted by la_lo626
design a cylinder which can hold a volume of 10ft cubed. so that it's used with minimal material. what are the dimensions (r and h) and what is the relationship between r and h. HINT: MINIMIZE SURFACE AREA????????????

Here is the basic set up.

The Volume of the cylinder is $V=\pi r^2h$

If the cylinder is closed (like a can of soup) the surface area will be the area of the circles on the top and bottom and the area of the side of the cylinder.

if you cut open a cylinder you would get two circles and one rectangle.

Now The big question is how long is the rectangle? It's length must be the same as the circumference of the circle. So its area will be

$A_{rec}=2\pi r h$ and the area of the circle is

$A_{circ}=\pi r^2$

We can use these to find the surface area of the can.

I'm not sure if your cylinder is open on top or not, but the process for both is similar.

So lets suppose that it is open on top then the surface area is

$A_{surface}=2\pi r h+\pi r^2$

but we also know that $10=\pi r^2 h \implies h= \frac{10}{\pi r^2}$

Now we can sub this into the surface area equation to get

$A=2 \pi r\left( \frac{10}{\pi r^2}\right)+\pi r^2=\frac{20}{r}+\pi r^2$

You should be able to finish from here.

Good luck
• November 23rd 2008, 06:58 PM
Soroban
Hello, la_lo626!

Quote:

Design a cylinder which can hold a volume of 10 ft³
so that its surface area is a minimum.

(a)What are the dimensions, $r$ and $h$?

The volume is 10 ft³: . $\pi r^2h \:=\:10 \quad\Rightarrow\quad h \:=\:\frac{10}{\pi r^2}$ .[1]

The surface area is: . $\begin{Bmatrix} \text{top/bottom:} & 2\!\times\! \pi r^2 \\ \text{lateral area:} & 2\pi rh\end{Bmatrix} \quad\Rightarrow\quad A \;=\;2\pi r^2 + 2\pi rh$ .[2]

Substitute [1] into [2]: . $A \;=\;2\pi r^2 + 2\pi r \left(\frac{10}{\pi r^2}\right)\quad\Rightarrow\quad A \:=\:2\pi r^2 + 20r^{-1}$

Differentiate and equation to zero: . $A' \;=\;4\pi r - 20r^{-2} \;=\;0$

Multiply by $r^2\!:\;\;4\pi r^3 -20 \:=\:0 \quad\Rightarrow\quad r^3 \:=\:\frac{5}{\pi} \quad\Rightarrow\quad\boxed{ r \:=\:\sqrt[3]{\frac{5}{\pi}}}$

$\text{Substitute into {\color{blue}[1]}: }\;h \;=\;\frac{10}{\pi\left(\sqrt[3]{\frac{5}{\pi}}\right)^2} \quad\Rightarrow\quad\boxed{ h \;=\;2\sqrt[3]{\frac{5}{\pi}}}$

Quote:

(b) What is the relationship between $r$ and $h$ ?
. . $h \;=\;2r$

Ha . . . $\emptyset$ beat me to it . . .
.