First slide

Maxima and minima

Question

If $θ$ is the angle (semi-vertical) of a cone of maximum volume and given slant height, then $\tan θ$ is given by

Moderate

A

2
B

1
C

$\sqrt{2}$
D

$\sqrt{3}$

Solution

$Let OB = 1, OA = 1 \cos θ and A B = 1 \sin θ (0 \leq θ \leq π / 2) . Then$
$\begin{aligned} V = \frac{π}{3} (A B)^{2} (O A) = \frac{π}{3} l^{3} \sin^{2} θ \cos θ \\ \Rightarrow \frac{d V}{d θ} = \frac{π}{3} l^{3} \sin θ (3 \cos^{2} θ - 1) \end{aligned}$
$\begin{array}{l} So from d V / d θ = 0, we get θ = 0 or \cos θ = 1 / \sqrt{3} . \\ Also, V (0) = 0, V (π / 2) = 0 \end{array}$
$and V (\cos^{- 1} \frac{1}{\sqrt{3}}) = \frac{2 π l^{3}}{9 \sqrt{3}}$
$Hence V is maximum when \cos θ = 1 / \sqrt{3}, i.e. \tan θ = \sqrt{2}$

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