Answer and Step-by-step explanation:
Solution:
Given:
V = 2π cm3
We know that:
2π = πr2h
2π / πr2 = h
Which gives: h = 2 / r2
Minimize surface area = 2πrh + 2πr2
S = 2πrh + 2πr2
Put h = 2/ r2 in surface area, we get:
S = 2πr (2/ r2 ) + 2πr2
S = 4 π / r + 2πr2
Therefore, s(r) = 4 π / r + 2πr2
Differentiate concerning r, we have
S’(r) = -4π / r2 + 4πr
Again differentiate concerning r, we have:
S”(r) = 2 x 4 π/ r3 + 4π
For s’(r) = 0
0 = -4π / r2 + 4πr
-4π + 4πr3 = 0
r3 = 1
r = 1 cm
and s”(1) > 0 at r = 1 is point of minima.
So h = 2 π / π x1 x1
= 2cm
Therefore minimum surface area the can should have a radius of 1cm and height of 2cm.