Question

Find the radius of the container in the shape of a right circular cylinder with an open top whose volume is 1 liters and whose surface area is minimal. Carefully explain why your answer would give the minimal surface area.

Detailed and step by step solution please :)

Answer 1

Give, Volume of right cylinder = 1 Ltr = 0.001 m^3

pi*r^2*h = 0.001

h = 0.001/(pi*r^2)    eq(1)

here, r = radius of cylinder

h = height of cylinder

now, surface area of open top cyliner(S) = 2*pi*r*h + pi*r^2

from eq(1),

S = (2*pi*r)*(0.001/(pi*r^2)) + pi*r^2

S = 0.002/(r) + pi*r^2

Now surface will be minimum at critical points of the function, So for critical point,

dS/dr = 0

S' = dS/dr = -0.002/r^2 + 2*pi*r = 0

pi*r = 0.001/r^2

r^3 = 0.001/pi

r = (0.001/pi)^(1/3)

r = 0.0683 m

Now to check whether this point will be minimum or maximum, Using 2nd derivative test:

S'' = 0.004/r^3 + 2*pi*1

S'' = 0.004/r^3 + 2*pi

At r = 0.0683, S'' > 0, So at this point surface area will be minimum

So radius of container = 0.0683 m

If you need height, then h = 0.001/(pi*0.0683^2) = 0.0683

Let me know if you've any query.