If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3cm, then the cu

Q: Wait, why is the distance from the center to the base $h - R$?

Great question! Think about the geometry. The total height of the cone is $h$. The distance from the apex of the cone to the center of the sphere is $R$. Therefore, the distance from the center of the sphere to the base of the cone must be the total height minus the radius of the sphere, which is $h - R$. If the base of the cone is below the center, this value becomes negative, but the square $(h - R)^2$ remains the same, so the math holds up perfectly!

Q: Why can't we just use energy conservation or some other shortcut?

This is a classic optimization problem, not a dynamics problem! We are looking for the geometric configuration that maximizes a specific property (volume). In such cases, calculus is our most powerful tool. We define the volume as a function of one variable and find where its rate of change is zero. There is no shortcut for the elegance of finding the critical point!

Q: Why do we need the slant height $l$ for the curved surface area?

The formula for the curved surface area of a cone is $\pi r l$, where $l$ is the slant height. The slant height is the hypotenuse of the right triangle formed by the height $h$ and the base radius $r$. Without $l$, you would only have the base area, not the surface area of the cone's side. Always remember to visualize the cone's net!

Question

If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3cm, then the curved surface area (in cm$^2$) of this cone is :

Options:

(A) $6 \sqrt{3} \pi$

(B) $6 \sqrt{2} \pi$

(C) $8 \sqrt{2} \pi$

(D) $8 \sqrt{3} \pi$

Sigma Percentile · Accepted Answer

### The Geometry of Perfection: Maximizing the Cone

Imagine you are an architect of the microscopic, tasked with fitting the most voluminous cone possible inside a sphere of radius $R=3$ cm. This is not just a math problem; it is a quest for the perfect balance between height and width.

When we inscribe a cone in a sphere, we are essentially playing a game of constraints. The sphere is our boundary, and the cone is our creation. Let us embark on this journey to find the curved surface area of this optimal cone.

### Phase 1: The Geometric Blueprint

To solve this, we must first visualize the cross-section. Imagine slicing the sphere right through the middle, passing through the apex of the cone. You see a circle with a triangle inside it.

The cone's height is $h$, and its base radius is $r$. The center of the sphere is our anchor. The distance from the apex to the center is $R$. Thus, the distance from the center to the base of the cone is $|h - R|$.

By applying the Pythagorean theorem to the right-angled triangle formed by the sphere's center, the base radius $r$, and the vertical distance $|h - R|$, we get the fundamental constraint:

$$r^2 + (h - R)^2 = R^2$$

Expanding this, we find $r^2 + h^2 - 2hR + R^2 = R^2$. The $R^2$ terms cancel out, leaving us with the elegant relationship:

$$r^2 = 2hR - h^2$$

This is the key that unlocks the entire problem.

### Phase 2: The Calculus of Optimization

The volume of a cone is given by the formula:

$$V = \frac{1}{3} \pi r^2 h$$

We have two variables, $r$ and $h$, but we want to maximize $V$. Using our constraint $r^2 = 2hR - h^2$, we can rewrite the volume entirely in terms of $h$:

$$V(h) = \frac{1}{3} \pi (2hR - h^2)h = \frac{\pi}{3} (2Rh^2 - h^3)$$

Now, we are in the realm of calculus. To find the maximum volume, we take the derivative with respect to $h$ and set it to zero:

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

Factoring this, we get $h(4R - 3h) = 0$. Since the height cannot be zero, we find the optimal height:

$$h = \frac{4R}{3}$$

With $R=3$, our optimal height is $h = 4$ cm.

### Phase 3: The Final Calculation

Now that we have the height, we find the base radius $r$. Substituting $h=4$ and $R=3$ into $r^2 = 2hR - h^2$, we get:

$$r^2 = 2(4)(3) - 16 = 24 - 16 = 8$$

Thus, $r = 2\sqrt{2}$ cm. To find the curved surface area, we need the slant height $l$:

$$l = \sqrt{r^2 + h^2} = \sqrt{8 + 16} = \sqrt{24} = 2\sqrt{6} 	ext{ cm}$$

Finally, the curved surface area is:

$$CSA = \pi r l = \pi (2\sqrt{2})(2\sqrt{6}) = 4\pi \sqrt{12} = 8\sqrt{3}\pi 	ext{ cm}^2$$

We have arrived at the solution, not by guessing, but by following the logical path of geometry and calculus. You have successfully optimized the cone, resulting in a final curved surface area of **$8\sqrt{3}\pi$ cm$^2$**.

Animated Solution for Mathematics - Differentiation: If a right circular cone, having maximum volume, is inscribed in a sphere of radius 3cm, then the curved surface area (in cm $^{2}$ ) of this cone is :

Select Answer:

Visualized Solution

Visualize the Geometry

Relate $r$ and $h$ using Pythagoras

Simplify the Relation

Define Volume Function $V (h)$

Simplify $V (h)$

Maximize Volume using Differentiation

Differentiate $V (h)$

Solve for Optimal Height $h$

Calculate Dimensions for $R = 3$

Calculate Base Radius $r$

Find Slant Height $l$

Calculate Slant Height $l$

Final Calculation of CSA

Simplify the Final Answer

The Sigma Insight: Maxima and Minima

The Geometry of Perfection

Phase 1

Phase 2

Phase 3

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