The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is (1) 2

Q: Wait, why is the height of the triangle $\frac{h}{2}$ and not $h$?

That is a fantastic question! Remember, the cylinder is inscribed symmetrically within the sphere. The center of the sphere acts as the midpoint of the cylinder's central axis. Therefore, the distance from the sphere's center to the top base of the cylinder is exactly half the total height, which is $\frac{h}{2}$. If you used $h$, you would be looking at the distance to the bottom base, which is incorrect for the Pythagorean triangle.

Q: Why do we use $a^2$ instead of $a$ in the volume formula?

It is a strategic choice! The volume of a cylinder is $V = \pi a^2 h$. Our Pythagorean relation gives us $a^2 = R^2 - \frac{h^2}{4}$ directly. By keeping $a^2$ as a single term, we avoid dealing with square roots, which makes the differentiation process much cleaner and less prone to algebraic errors.

Q: How do we know this is a maximum and not a minimum?

Great intuition! To be absolutely sure, we can look at the second derivative. The first derivative is $\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4}$. The second derivative is $\frac{d^2V}{dh^2} = -\frac{6\pi h}{4} = -\frac{3\pi h}{2}$. Since $h$ must be positive, the second derivative is negative, which confirms that our critical point is indeed a local maximum.

Question

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is

Options:

(A) $2\sqrt{3}$

(B) $\sqrt{3}$

(C) $\sqrt{6}$

(D) $2/3 \sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of Perfection: Maximizing Volume in a Sphere

Welcome, fellow explorer of physics and mathematics. Today, we are not just solving a problem; we are embarking on a journey to find the perfect balance.

Imagine you are an architect of the microscopic world, tasked with fitting the largest possible cylinder inside a sphere of radius $R = 3$. This is a classic JEE Advanced problem, and it is a beautiful example of how geometry and calculus dance together to reveal the optimal state of a system.

### Phase 1: Visualizing the Geometry

Before we touch a single equation, let us close our eyes and visualize. We have a sphere—a perfect, symmetric object. Inside, we want to place a cylinder.

If we slice this sphere right down the middle, through the center and along the axis of the cylinder, what do we see? We see a rectangle inscribed within a circle.

The circle has a radius $R$, and the rectangle has a width of $2a$ (where $a$ is the cylinder's radius) and a height of $h$. The center of the sphere is the center of this rectangle. This visualization is the key to everything.

### Phase 2: The Pythagorean Bridge

Now, let us connect the sphere's dimensions to the cylinder's dimensions. Look at the right-angled triangle formed by the center of the sphere, the edge of the cylinder's base, and the center of the cylinder's base.

The hypotenuse of this triangle is the sphere's radius, $R$. The base of this triangle is the cylinder's radius, $a$. The height of this triangle is half the cylinder's height, $\frac{h}{2}$.

By the Pythagorean theorem, we have the fundamental relationship:

$$a^2 + \left( \frac{h}{2} \right)^2 = R^2$$

This is our bridge. It allows us to express the cylinder's radius in terms of its height:

$$a^2 = R^2 - \frac{h^2}{4}$$

### Phase 3: The Volume Function

We want to maximize the volume $V$ of the cylinder. The formula for the volume of a cylinder is $V = \pi a^2 h$.

Notice that we have two variables here: $a$ and $h$. But we just found a way to express $a^2$ in terms of $h$. Let us substitute that into our volume formula:

$$V = \pi \left( R^2 - \frac{h^2}{4} \right) h$$

Now, our volume is a function of a single variable, $h$:

$$V(h) = \pi R^2 h - \frac{\pi h^3}{4}$$

This is the function we need to maximize. It is a simple polynomial, and we know exactly how to handle it.

### Phase 4: The Calculus of Optimization

To find the maximum, we need to find the critical points. We take the derivative of $V$ with respect to $h$ and set it to zero:

$$\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4} = 0$$

This is the moment of truth. Solving for $h^2$, we get:

$$\frac{3\pi h^2}{4} = \pi R^2 \Rightarrow h^2 = \frac{4R^2}{3}$$

Taking the square root, we find the optimal height:

$$h = \frac{2R}{\sqrt{3}}$$

This is a universal result! For any sphere of radius $R$, the inscribed cylinder of maximum volume will always have this height.

### Phase 5: The Final Calculation

Finally, let us apply this to our specific problem where $R = 3$. Substituting $R = 3$ into our formula, we get:

$$h = \frac{2(3)}{\sqrt{3}} = \frac{6}{\sqrt{3}}$$

To make this look elegant, we rationalize the denominator:

$$h = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$$

And there it is! The height of the cylinder that maximizes the volume is **$2\sqrt{3}$**.

### Conclusion

Mathematics is not just about getting the right answer; it is about understanding the elegance of the process. We started with a sphere and a cylinder, used the Pythagorean theorem to link them, built a volume function, and used calculus to find the peak.

You have mastered the geometry, the algebra, and the calculus. Keep this mindset, and no problem will ever be too complex for you. You are doing great!

Animated Solution for Mathematics - Differentiation: The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is

Select Answer:

Visualized Solution

Visualizing the Setup

Extracting the Geometry

Applying Pythagoras Theorem

Isolating $a^{2}$

Volume of a Cylinder

Single Variable Volume Function

Expanding the Expression

Maximizing the Volume

Differentiating $V$

Equating Derivative to Zero

Rearranging for $h^{2}$

General Formula for Optimal Height

Substituting the Given Radius

Final Calculation

The Sigma Insight: Maxima and Minima

The Geometry of Perfection

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Conclusion

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