Q.Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $\tan^{-1}\sqrt{2}$.

Concept: Optimization Word Problem — maximizing volume of a cone under a fixed slant height constraint.

Let the cone have slant height $l$ (fixed), radius $r$, and height $h$. By Pythagoras, $r^2 + h^2 = l^2$. Volume is $V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (l^2 - h^2)h$.

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

$\frac{dV}{dh} = \frac{1}{3}\pi (l^2 - 3h^2) = 0 \implies h = \frac{l}{\sqrt{3}}$. …

We have a cone with a fixed slant height $l$. The slant height is the distance from the apex to any point on the circular base, measured along the sloping surface. The semi-vertical angle $\theta$ is the angle between the axis (height) and the slant height. Our job: find $\theta$ that gives the largest possible volume.

Why does this work? In optimisation problems with a constraint (here, fixed $l$), we express the quantity to be maximised — volume — in terms of a single variable. The slant height ties the radius $r$ and height $h$ together via $l^2 = r^2 + h^2$. So we can write $r$ and $h$ in terms of $\theta$ and $l$, then volume becomes a function of $\theta$ alone. Differentiate, set to zero, and check it's a maximum.

Let's go step by step.

1. Relate the cone's dimensions to $\theta$ and $l$

In a right circular cone, the semi-vertical angle $\theta$ is at the apex, between the axis (height $h$) and the slant height $l$. So:

• $\sin \theta = \frac{r}{l}$ → $r = l \sin \theta$
• $\cos \theta = \frac{h}{l}$ → $h = l \cos \theta$

These follow directly from the right triangle formed by $r$, $h$, and $l$.

2. Write the volume in terms of $\theta$

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

Substitute $r$ and $h$:

Since $l$ is constant, maximising $V$ is equivalent to maximising $f(\theta) = \sin^2 \theta \cos \theta$.

3. Differentiate $f(\theta)$ and set to zero

Let $f(\theta) = \sin^2 \theta \cos \theta$. Use the product rule:

Simplify:

Factor out $\sin \theta$:

For a maximum in $(0, \frac{\pi}{2})$, set $f'(\theta) = 0$. $\sin \theta \neq 0$ (since $\theta > 0$), so:

4. Solve for $\theta$ …

Method: Optimizing via an Angle Parameter When the Slant Height Is Fixed

When a cone problem fixes the slant height $l$ and asks about the semi-vertical angle $\theta$, it is usually cleanest to express everything in terms of $\theta$ rather than $r$ or $h$ directly, because $\theta$ is a single independent variable once $l$ is fixed.

Steps

Step 1: Express $r$ and $h$ in terms of $\theta$ and the fixed $l$

In the right triangle formed by the axis, the base radius, and the slant side:

Keep $\sin$ paired with the base radius and $\cos$ paired with the height — mixing these up is the single most common error in this method.

Step 2: Write the quantity to optimize (volume) in terms of $\theta$ alone

Since $l$ is a fixed constant, you may drop the constant factor $\dfrac{1}{3}\pi l^3$ and just maximize $f(\theta) = \sin^2\theta\cos\theta$ — this location of the maximum is unchanged, but the algebra becomes much lighter.

Step 3: Differentiate using the product rule and factor

$f'(\theta) = 2\sin\theta\cos^2\theta - \sin^3\theta = \sin\theta\left(2\cos^2\theta - \sin^2\theta\right)$ …

Common Mistakes

Mistake 1: Swapping $\sin\theta = r/l$ and $\cos\theta = h/l$

Because the semi-vertical angle is measured from the axis, it is $\cos\theta$ that pairs with the height and $\sin\theta$ that pairs with the base radius. Reversing these two relations flips the final answer (you would get $\tan\theta = \dfrac{1}{\sqrt2}$ instead of $\sqrt2$).

Mistake 2: Trying to differentiate the full expression with the constant $\dfrac{1}{3}\pi l^3$ still attached …