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Q26:
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is $\sin^{-1}(\frac{1}{3})$.

Solution :

Step 1: Establishing the Geometric Model

Let the cone have radius $r$, slant height $l$, vertical height $h$, and semi-vertical angle $\alpha$.

  • $h = l \cos \alpha$ [By Trigonometric definition in $\triangle AOB$]
  • $r = l \sin \alpha$
  • Surface Area $S = \pi r^2 + \pi rl$ (Total Surface Area)
h r $\alpha$

Step 2: Expressing $l$ and $h$ in terms of $r$ and $\alpha$

Given $S = \pi r^2 + \pi r(\frac{r}{\sin \alpha}) = \pi r^2(1 + \frac{1}{\sin \alpha})$.

Solving for $r^2$: $r^2 = \frac{S}{\pi(1 + \csc \alpha)}$.

The volume $V$ is given by $V = \frac{1}{3}\pi r^2 h$. Substituting $h = r \cot \alpha$:

$V = \frac{1}{3}\pi r^2 (r \cot \alpha) = \frac{1}{3}\pi r^3 \cot \alpha$.

Step 3: Optimization via Derivative

To maximize $V$, it is convenient to maximize $V^2$ (as $V>0$):

$V^2 = \frac{\pi^2}{9} r^6 \cot^2 \alpha = \frac{\pi^2}{9} \left( \frac{S}{\pi(1 + \csc \alpha)} \right)^3 \cot^2 \alpha$.

Let $f(\alpha) = \frac{\cot^2 \alpha}{(1 + \csc \alpha)^3} = \frac{\frac{\cos^2 \alpha}{\sin^2 \alpha}}{(\frac{\sin \alpha + 1}{\sin \alpha})^3} = \frac{\cos^2 \alpha \sin \alpha}{(1 + \sin \alpha)^3} = \frac{(1 - \sin^2 \alpha) \sin \alpha}{(1 + \sin \alpha)^3} = \frac{(1 - \sin \alpha)(1 + \sin \alpha) \sin \alpha}{(1 + \sin \alpha)^3}$.

$f(\alpha) = \frac{\sin \alpha - \sin^2 \alpha}{(1 + \sin \alpha)^2}$.

Step 4: Solving $f'(\alpha) = 0$

Let $x = \sin \alpha$. Then $f(x) = \frac{x - x^2}{(1 + x)^2}$.

Using the Quotient Rule $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$:

$f'(x) = \frac{(1 - 2x)(1 + x)^2 - (x - x^2) \cdot 2(1 + x)}{(1 + x)^4} = 0$.

$(1 - 2x)(1 + x) - 2(x - x^2) = 0$.

$1 + x - 2x - 2x^2 - 2x + 2x^2 = 0$.

$1 - 3x = 0 \implies x = \frac{1}{3}$.

Step 5: Conclusion

Since $x = \sin \alpha = \frac{1}{3}$, we have $\alpha = \sin^{-1}(\frac{1}{3})$.

By the Second Derivative Test, since the volume is bounded and we found a unique critical point, this corresponds to the local maximum [Per First Derivative Test/Global Extremum Theorem].

Final Solution: The semi-vertical angle is $\alpha = \sin^{-1}(\frac{1}{3})$.


More Questions from Class 12 Mathematics Application of Derivatives EXERCISE 6.3


CBSE Solutions for Class 12 Mathematics Application of Derivatives


Chapters in CBSE - Class 12 Mathematics


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