First slide

Maxima and minima

Question

If $A > 0, B > 0, and A + B = \frac{π}{3},$ then the maximum value of tan A tan B is

Moderate

A

$\frac{1}{\sqrt{3}}$
B

$\frac{1}{3}$
C

3
D

$\sqrt{3}$

Solution

Given $A + B = 60^{\circ} or B = 60^{\circ} - A$
$∴ \tan B = \tan (60^{\circ} - A) = \frac{\sqrt{3} - \tan A}{1 + \sqrt{3} \tan A}$
$Now z = \tan Atan B$
or $z = \frac{t (\sqrt{3} - t)}{1 + \sqrt{3} t} = \frac{\sqrt{3} t - t^{2}}{1 + \sqrt{3} t}$
where t = tan A
$\frac{dz}{dt} = - \frac{(t + \sqrt{3}) (\sqrt{3} t - 1)}{(1 + \sqrt{3} t)^{2}} = 0$
or $t = 1 / \sqrt{3}$
or $t = \tan A = \tan 30^{\circ}$
The other value is rejected as both A and B are +ve acute angles.
If $t < \frac{1}{\sqrt{3}}, \frac{dz}{dt}$ is positive and if $t > \frac{1}{\sqrt{3}}, \frac{dz}{dt}$ is negative.
Hence, maximum when $t = \frac{1}{\sqrt{3}}$ and maximum value $= \frac{1}{3} .$

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