The moment of inertia of a uniform cylinder of length l and radius R about its perpendicular bisector is I. What is the ratio l/R such that the moment of inertia is minimum?

Given, The length of the cylinder = l, Radius of cylinder = R And I = moment of inertia.
Now, I = mR²/4 + ml²/12
I = m/4(R² + l²/3)
I = m/4(Vπl + l²/3) (∵V = πR²l)
Now differentiate I with respect to l
So, dI/dl = m/4(-Vπl² + 2l/3)
For maxima and minima, dI/dl = 0
So, m/4(-Vπl² + 2l/3) = 0
⇒ Vπl² = 2l/3
R²l = 2l/3 (∵V = πR²l)
⇒ l²/R² = 3/2
⇒ √(l²/R²) = √(3/2)
l/R = √(3/2)