A cylindrical container is to be made from certain solid material with the following constraints: It

Q: Wait, why is the outer height $H = h + 2$? Shouldn't it just be $h$?

That is a classic trap! Remember, the problem states the bottom of the container is a solid circular disc of thickness $2$ mm. If the inner height is $h$, the total height of the container must include that bottom base. So, $H = h + 2$. Always sketch the cross-section to avoid these geometric pitfalls.

Q: Can I use Lagrange Multipliers to solve this instead of substitution?

Absolutely! Lagrange Multipliers are a powerful tool for constrained optimization. You would define the Lagrangian $\mathcal{L}(r, h, \lambda) = M(r, h) + \lambda(\pi r^2 h - V)$. While it is mathematically elegant, for this specific problem, direct substitution is often faster and less prone to algebraic errors. Stick to the method that keeps your mind clear.

Q: Why do we ignore the $-V$ term when differentiating? Doesn't it affect the minimum?

Great observation! When we differentiate $M$ with respect to $r$, we are looking for the rate of change. Since $V$ is a fixed constant (the inner volume is given as fixed), its derivative with respect to $r$ is zero. It shifts the graph up or down, but it does not change the location of the minimum point. Keep your focus on the variable terms!

Question

A cylindrical container is to be made from certain solid material with the following constraints: It has a fixed inner volume of $V$ mm$^3$, has a $2$ mm thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness $2$ mm and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is $10$ mm, then the value of $V / 250\pi$ is

Sigma Percentile · Accepted Answer

### The Engineer's Dilemma: Optimizing the Vessel

Welcome, future engineers and physicists. Today, we are not just solving a calculus problem; we are stepping into the shoes of a design engineer.

Imagine you are tasked with creating a cylindrical container. You have a specific requirement: it must hold a fixed volume $V$ of liquid. But here is the catch—you want to use the absolute minimum amount of material to build it.

This is a classic optimization problem, the kind that separates the casual student from the master of JEE Advanced. Let us break this down, layer by layer.

### Phase 1: Visualizing the Geometry

The first step in any geometry problem is to stop looking at the numbers and start looking at the object. We have an inner cylinder where the liquid sits. Let its radius be $r$ and its height be $h$.

The problem tells us the walls are $2$ mm thick and the bottom is $2$ mm thick. If you are standing inside the container, the walls are $2$ mm away from you. If you are looking at the container from the outside, the radius has expanded by that thickness.

So, the outer radius $R$ is $r + 2$. Now, look at the height. The inner height is $h$. But there is a solid base of $2$ mm at the bottom.

Therefore, the total outer height $H$ is $h + 2$. This is where most students stumble. They forget the base thickness!

Always draw the cross-section. Once you see that $R = r + 2$ and $H = h + 2$, the geometry becomes your ally, not your enemy.

### Phase 2: The Constraint of Volume

We are given that the inner volume $V$ is fixed. We know the formula for the volume of a cylinder is $V = \pi r^2 h$.

Since $V$ is a constant, we have a beautiful relationship between $h$ and $r$:

$$h = \frac{V}{\pi r^2}$$

This equation is our key. It allows us to eliminate the variable $h$ later on. Whenever you see two variables in an optimization problem, look for a constraint equation like this. It is the bridge that connects your variables.

### Phase 3: Defining the Material Volume

Now, we need to define the function we want to minimize: the volume of the material, $M$. The material volume is simply the total volume of the cylinder minus the empty space inside.

$$M = 	ext{Total Volume} - 	ext{Inner Volume}$$

Using our outer dimensions, the total volume is $\pi R^2 H$. Substituting our expressions for $R$ and $H$, we get:

$$M = \pi (r+2)^2 (h+2) - V$$

This looks a bit intimidating, doesn't it? But do not panic. We have our constraint $h = \frac{V}{\pi r^2}$. Let us substitute that in:

$$M = \pi (r+2)^2 \left(\frac{V}{\pi r^2} + 2ight) - V$$

### Phase 4: The Calculus of Optimization

Before we differentiate, let us simplify. If you try to use the product rule on that expression as it stands, you might get lost in the algebra. Let us distribute the terms:

$$M = \pi (r+2)^2 \frac{V}{\pi r^2} + 2\pi(r+2)^2 - V$$

Notice the magic? The $\pi$ cancels out in the first term! We are left with:

$$M = V \left(1 + \frac{2}{r}ight)^2 + 2\pi(r+2)^2 - V$$

Now, we differentiate with respect to $r$. We are looking for the minimum, so we set $\frac{dM}{dr} = 0$. Remember, $V$ is a constant, so its derivative is zero. Applying the chain rule:

$$\frac{dM}{dr} = V \cdot 2\left(1 + \frac{2}{r}ight) \cdot \left(-\frac{2}{r^2}ight) + 2\pi \cdot 2(r+2) = 0$$

### Phase 5: The Final Calculation

The problem tells us the minimum occurs at $r = 10$ mm. This is our golden ticket. We substitute $r = 10$ into our derivative equation:

$$V \cdot 2\left(1 + \frac{2}{10}ight) \cdot \left(-\frac{2}{100}ight) + 4\pi(10+2) = 0$$

Let us simplify the arithmetic carefully:

$$V \cdot 2(1.2) \cdot (-0.02) + 48\pi = 0$$

$$-0.048 V + 48\pi = 0$$

Solving for $V$, we get:

$$0.048 V = 48\pi \implies V = \frac{48\pi}{0.048} = 1000\pi$$

Finally, the question asks for the value of $\frac{V}{250\pi}$. Substituting our value for $V$:

$$\frac{1000\pi}{250\pi} = 4$$

And there it is. The **answer is 4**. You have successfully navigated the geometry, the constraints, and the calculus. This is the essence of JEE Advanced physics and math—taking a complex, real-world scenario and distilling it into a beautiful, elegant solution.

Enter Numerical Value:

Visualized Solution

Visualizing the Container Geometry

Expressing Height in terms of Volume

Defining Material Volume $M$

Substituting $h$ into the Equation

Simplifying for Differentiation

Applying the Minimization Condition

Differentiating the Function

Substituting $r = 10$

Solving for Volume $V$

Final Answer Calculation

The Sigma Insight: Maxima and Minima

The Engineer's Dilemma

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

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