Q.Of all the closed cylindrical cans (right circular), of a given volume of $100$ cubic centimetres, find the dimensions of the can which has the minimum surface area?

Concept: Optimization Word Problem — minimizing surface area for a fixed volume.

Let the radius be $r$ cm and height $h$ cm. Volume is fixed:

$\pi r^2 h = 100 \quad\Rightarrow\quad h = \frac{100}{\pi r^2}$.

Surface area (including top and bottom) is:

$S = 2\pi r^2 + 2\pi r h = 2\pi r^2 + 2\pi r \cdot \frac{100}{\pi r^2} = 2\pi r^2 + \frac{200}{r}$.

Differentiate with respect to $r$ and set to zero:

$\frac{dS}{dr} = 4\pi r - \frac{200}{r^2} = 0 \quad\Rightarrow\quad 4\pi r^3 = 200 \quad\Rightarrow\quad r^3 = \frac{50}{\pi}$. …

This is a classic optimisation problem from calculus — but the real insight is geometric. You have a fixed volume to enclose, and you want to use as little material as possible. For a cylinder, the surface area is the sum of the curved side and two circular ends. The trick is that the ends cost "more" surface per unit of enclosed volume than the side does, so you want to make the cylinder taller and narrower — but not too tall, because then the side area grows. The balance point is where the height equals the diameter.

Let’s work it out.

1. Set up the variables and constraints. Let the radius be $r$ cm and the height be $h$ cm. The volume is fixed:

The surface area (top, bottom, and curved side) is:

2. Eliminate one variable using the constraint. From $V = 100$, we have $h = \frac{100}{\pi r^2}$. Substitute into $S$:

Now $S$ is a function of $r$ alone. The domain is $r > 0$.

3. Differentiate to find the critical point.

Set $S'(r) = 0$:

So the critical radius is:

4. Confirm it’s a minimum.

   The second derivative is $S''(r) = 4\pi + \frac{400}{r^3}$, which is positive for all $r > 0$. Hence the function is convex, and this critical point gives a global minimum.

5. Find the corresponding height. …

Method: Minimizing Surface Area of a Solid for a Fixed Volume

This method applies whenever a problem fixes the volume of a solid (a can, a box, a container) and asks you to make its surface area as small as possible — the classic "least material for a given capacity" problem.

Steps

Step 1: Write both quantities — the objective and the constraint — in terms of the solid's natural variables

For a closed right circular cylinder of radius $r$ and height $h$:

Always include every surface the solid actually has — a "closed" can has a top and a bottom, so both circular ends belong in $S$.

Step 2: Use the constraint to eliminate one variable

Since $V$ is a fixed number, solve $V = \pi r^2 h$ for $h$:

Substitute this into $S$ so that $S$ becomes a function of the single variable $r$:

Step 3: Differentiate and find the critical point

Set $S'(r) = 0$ and solve for $r$ — this gives the radius at which $S$ is stationary. …

Common Mistakes

Mistake 1: Leaving out one of the circular ends

A "closed" cylindrical can has two circular ends (top and bottom), not one. Writing $S = 2\pi r^2 + 2\pi rh$ with only a single $\pi r^2$ term (as if it were an open can) is wrong — it changes the optimum ratio from $h = 2r$ to $h = r$. Read "closed" carefully before setting up $S$.

Mistake 2: Stopping after solving $S'(r) = 0$ without confirming a minimum …