# Least Surface Area Problem

• April 13th 2010, 10:50 AM
WartonMorton
Least Surface Area Problem
• April 13th 2010, 10:52 AM
earboth
Quote:

Originally Posted by WartonMorton

I can't see any attachment (Crying)
• April 13th 2010, 11:03 AM
WartonMorton
Quote:

Originally Posted by earboth
I can't see any attachment (Crying)

Sorry about that. I think I figured out how to attach it now.
• April 13th 2010, 11:16 AM
running-gag
Hi

Let R be the radius of the cylinder and L its length

The volume is $V_0 = \pi R^2L + \frac43 \pi R^3 = 13.4 cm^3$

From this expression you can find L as a function of R

The area is $A(R,L) = 2\pi RL + 4 \pi R^2$

Substitute L function of R to get A(R)

Calculate A'(R) to find the value of R which minimizes the area
• April 13th 2010, 11:20 AM
WartonMorton
Quote:

Originally Posted by running-gag
Hi

Let R be the radius of the cylinder and L its length

The volume is $V_0 = \pi R^2L + \frac43 \pi R^3 = 13.4 cm^3$

From this expression you can find L as a function of R

The area is $A(R,L) = 2\pi RL + 4 \pi R^2$

Substitute L function of R to get A(R)

Calculate A'(R) to find the value of R which minimizes the area

I got 1.845, how does this match up to what you got?
• April 13th 2010, 11:31 AM
running-gag
I am getting 1.473 in (Wondering)
• April 13th 2010, 05:18 PM
WartonMorton
Quote:

Originally Posted by running-gag
I am getting 1.473 in (Wondering)

Any body else got 1.845 or 1.473?
• April 13th 2010, 10:44 PM
earboth
Quote:

Originally Posted by WartonMorton
Any body else got 1.845 or 1.473?

The exact value of r is: $r = \frac{\sqrt[3]{10050}}{10 \cdot \sqrt[3]{\pi}} \approx 1.473461286$