Q.Prove that the volume of the largest cone that can be inscribed in a sphere of radius $R$ is $\frac{8}{27}$ of the volume of the sphere.

Concept: Optimization Word Problem — maximize the volume of a cone inscribed in a fixed sphere.

Let the sphere have radius $R$. Place the cone with vertex at the top of the sphere and base a distance $h$ below the centre. The height of the cone is $R + h$, and the radius of its base is $\sqrt{R^2 - h^2}$ (from the right triangle with hypotenuse $R$).

Volume of the cone:

Differentiate $V$ with respect to $h$ and set to zero:

so $R^2 - 2hR - 3h^2 = 0$. Solving: $(R - 3h)(R + h) = 0$, giving $h = R/3$ (positive root).

Substitute $h = R/3$: …

This is a classic optimization problem: you have a fixed sphere, and you want to fit the biggest possible cone inside it. The cone must touch the sphere’s surface at its base circumference and at its apex. The key insight is that the cone’s dimensions are not independent — they are linked by the geometry of the sphere. Once you express the cone’s volume in terms of a single variable (like its height), you can use calculus to find the maximum.

Let’s work through it step by step.

1. Set up the geometry

Imagine a sphere of radius $R$. Place the cone so that its apex is at the top of the sphere and its base is a circle somewhere below. The axis of the cone passes through the centre of the sphere. Let the height of the cone be $h$, measured from the apex to the centre of the base. The centre of the sphere lies somewhere along this axis.

If the apex is at the top, the centre of the sphere is at a distance $R$ from the apex. The base of the cone is at a distance $h$ from the apex. So the distance from the sphere’s centre to the base plane is $h - R$ (since $h > R$ for a cone that fits inside).

2. Relate base radius to height

The base of the cone is a circle of radius $r$. This circle lies on the sphere’s surface, so every point on its circumference is at distance $R$ from the sphere’s centre. The centre of the base is at a distance $h - R$ from the sphere’s centre. By the Pythagorean theorem in the vertical cross-section:

So:

Thus:

This is valid only when $2hR - h^2 \ge 0$, i.e. $0 \le h \le 2R$. The cone’s height cannot exceed the sphere’s diameter.

3. Write the volume of the cone

Volume of a cone is $V = \frac{1}{3}\pi r^2 h$. Substitute $r^2$:

So:

We need to maximise this for $h$ in $(0, 2R)$.

4. Differentiate and find critical points

Differentiate with respect to $h$:

Set $V'(h) = 0$:

So $h = 0$ (minimum, degenerate cone) or $h = \frac{4R}{3}$.

Since $h = \frac{4R}{3}$ lies in $(0, 2R)$, it is a valid candidate.

5. Confirm it’s a maximum

Check the second derivative:

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

So it is indeed a maximum. …

Method: Optimizing a Solid Inscribed Inside a Fixed Solid

This method applies whenever one solid (here, a cone) is inscribed inside another fixed solid (a sphere), and every point of the inscribed solid's boundary that touches the outer solid must satisfy the outer solid's own equation.

Steps

Step 1: Set up a clear geometric picture and label one variable along the axis

Place the sphere of fixed radius $R$ with its centre on the cone's axis. Let $h$ be a distance measured along that axis (for example, from the sphere's centre to the cone's base) — this single length will parametrize every inscribed cone.

Step 2: Use the sphere's own equation (Pythagoras) to relate the cone's base radius to $h$

Every point on the circular rim of the cone's base lies on the sphere, so it is at distance $R$ from the centre. This gives a right triangle linking the base radius $r$, the offset $h$, and $R$:

Solve this for $r^2$ in terms of $h$ and $R$ — this is the key geometric constraint that makes the problem one-variable.

Step 3: Write the cone's volume as a function of the single variable

Substitute the $r^2$ expression from Step 2 so $V$ depends only on $h$. …

Common Mistakes

Mistake 1: Mis-relating the cone's height to the offset from the sphere's centre

It is tempting to assume the cone's height equals the sphere's radius $R$, but geometrically the apex-to-base height is $R + h$ (or $R-h$, depending on how the offset is measured) once the base is offset from the centre. Skipping this careful setup and using $h = R$ outright gives a cone that is not actually the largest one.

Mistake 2: Forgetting the valid domain for the axial variable

The relation $r^2 = 2hR - h^2$ (or similar) is only valid while $r^2 \ge 0$, which restricts $h$ to a bounded interval. A critical point found without checking it lies strictly inside this interval — and without checking that the volume genuinely drops to $0$ at both ends — cannot be confirmed as the true maximum. …