The document describes minimizing the surface area of a cylindrical container with a volume of 250 ml. The variables of radius (r) and height (h) are defined. Taking the derivative of the surface area formula and setting it equal to 0 provides an expression for r in terms of the volume. Plugging this back into the original volume equation provides an expression for h in terms of r. The results are an optimal radius of 3.41 cm and height of 6.83 cm, minimizing the surface area.

2. Problem:
A cylindrical container is to have a
volume of 250ml. Find the
dimensions of the container that
minimise its surface area (Assume
the container is closed at both
ends).
2r
h

3. Solution:
Variables:
2 3
2 2
2 2 1
250 250
250
. ( ) 2 2 2 500
V r h ml cm
V
h cm cm
r r
S A f r r rh r r

 
   
  
  
    

4. Solution:
Differentiation:
2
2
3 3
( . ) ( ( ))
4 500
( . )
At Stationary point, 0, 4 500 0
500
4 500
4
d S A d f r
r r
dr dr
d S A
r r
dr
r r






  
   
   

5. Solution:
Further differentiation:
3
2 2
3
2 2
2
2
500
4
3
( . ) ( ( ))
4 1000
( . )
At minimum point, 0
which is the case here!
500
. is mimum at
4
r
d S A d f r
r
dr dr
d S A
dr
S A r





  

 

6. Solution:
Plot:

7. Solution:
Answer:
3
2
Required cylinder dimensions:
500
3.41
4
250
6.83
r cm cm
h cm cm
r


 
 