Question

Of all the closed right circular cylindrical cans of volume 128π cm³, find the dimensions of the can which has minimum surface area.

Answer 1

Let r and h be the radius and height of the cylindrical can respectively.
Therefore, the total surface area of the closed cylinder is given by

\[S = 2\pi r^2 + 2\pi rh\]        ...(1)

Given, volume of the can = 128π cm³

Also, volume (V) = \[\pi r^2 h\]

\[\therefore h = \frac{128}{r^2}\]                ... (2)

Putting the value of h in equation (1), we get:

\[S = 2\pi r^2 + 2\pi r \times \frac{128}{r^2}\]

\[\Rightarrow S = 2\pi r^2 + 2\pi \times \frac{128}{r}\]

Now, differentiating S with respect to r, we get:

\[\frac{dS}{dr} = 4\pi r - 2\pi \times \frac{128}{r^2}\]

Substituting 

\[\frac{dS}{dr} = 0\]  for the critical points, we get:

\[4\pi r - 2\pi \times \frac{128}{r^2} = 0\]

\[\Rightarrow r^3 = 64 \Rightarrow r = 4\]

Now, second derivative of S is given by

\[\frac{d^2 S}{d r^2} = 4\pi - 2\pi \times \left( - 2 \right)\frac{128}{r^3} = 4\pi + 4\pi \times \frac{128}{64} > 0 \left( \because r = 4 \right)\]

Thus, the total surface area of the cylinder is minimum when r = 4.
From equation (2), we have:

\[h = \frac{128}{4^2} = 8\]

Thus, the dimensions of the cylindrical can are r = 4 and h = 8.