Optimization Problem

**tradar** · April 23rd 2009, 02:08 PM

You have been asked to design a 1-liter cylindrical can made with sheet
aluminum. What dimensions of the can (radius and height) will use the least total
amount of aluminum? State dimensions to the nearest hundredth of a
centimeter.
Hint 1: The aluminum must cover both ends of the can as
well as the circular wall.
Hint 2: 1 liter = 1000 cm^3

I believe my primary equation would be: 2TTr^2 + 2TTrh

I think my secondary equation would be: TTr^2h=1000

Is that right?

I came up with a radius of 12.62 and a height of 2.00

We are supposed to have all of the following...
• Primary equation
• Secondary equation, if applicable
• Statement of domain
• Critical numbers and FDT to establish relative extrema
• Concluding statement(s) to answer the question.

**skeeter** · April 23rd 2009, 02:35 PM

$V = \pi r^2 h$

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

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

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

$A = 2 \pi r^2 + \frac{2V}{r}<br />$

$\frac{dA}{dr} = 4\pi r - \frac{2V}{r^2}$

set $\frac{dA}{dr} = 0$ ...

$4\pi r = \frac{2V}{r^2}$

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

$r = \left(\frac{V}{2\pi}\right)^{\frac{1}{3}}$

calculate $r$ ... go back and calculate $h$

you should see that $h = 2r$

**tradar** · April 23rd 2009, 02:43 PM

Ok, so the minimum surface area would be when the radius is 5.42 and the height is 10.84...right?

**Soroban** · April 23rd 2009, 02:58 PM

Hello, tradar!

Your answers are correct . . .

You are to design a 1-liter cylindrical can made with sheet aluminum.
What dimensions of the can (radius and height) will use the least total
amount of aluminum?
State dimensions to the nearest hundredth of a centimeter.

Hint 1: The aluminum must cover both ends of the can as well as the wall.
Hint 2: 1 liter = 1000 cm³.

I believe my primary equation would be: . $A \:=\: 2\pi r^2 + 2\pi rh$

I think my secondary equation would be: . $\pi r^2h\:=\:1000$

Is that right? . . . . Yes!

I came up with a radius of 12.62 and a height of 2.00 . . . . no

We have: . $\pi r^2h \:=\:1000 \quad\Rightarrow\quad h \:=\:\frac{1000}{\pi r^2}$ .[1]

Then: . $A\;=\;2\pi r^2 + 2\pi r\left(\frac{1000}{\pi r^2}\right) \;=\;2\pi r^2 + 2000r^{-1}$

Differentiate: . $A' \;=\;4\pi r - 2000r^{-2} \:=\:0 \quad\Rightarrow\quad 4\pi r^3 - 2000 \:=\:0$

. . $4\pi r^3 \:=\:2000 \quad\Rightarrow\quad r^3 \:=\:\frac{500}{\pi} \quad\Rightarrow\quad r \:=\:\sqrt[3]{\frac{500}{\pi}} \:=\:5\sqrt[3]{\frac{4}{\pi}}$

Substitute into [1] and we get: . $h \:=\:10\sqrt[3]{\frac{4}{\pi}}$

Therefore: . $r \:\approx\:5.42\text{ cm},\;\;h \:\approx\:10.84\text{ cm}$

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