Find the economical proportion between the radius and height of the cylindrical can to give the

Question

least dimensions of a metal that encloses a volume of 10 cu.

Answer 1

Let the radius of the can be r and its height be h
V = π r^2 h
10 = πr^2h or h = 10/(πr^2)
For your least dimension, the surface area (SA) has to be a minimum

SA = 2πr^2 + 2πrh
= 2πr^2 + 2πr(10/πr^2) = 2πr^2 + 20/r

D(SA)/dr = 4πr - 20/r^2 = 0 for a minimum of SA

4πr = 20/r^2
r^3 = 5/π = 1.591549
h = approx 1.1675
sub this back into h
h = approx 2.335
which is twice the value of r

So the height should be twice the radius
or
for a minimum SA , h : r = 2 : 1
(the height should equal the diameter)