Question

Show that ${\sin }^p\theta {\cos }^q\theta$ is maximum when $\theta ={\tan }^{-1}\sqrt{p/q}.$

Answer 1

$y={\sin }^p\theta {\cos }^q\theta$

Taking log on both sides

$\log y=p\log \sin \theta +q\log \cos \theta$

Now $y$ will be maximum or minimum according as $z=\log y=p\log \sin \theta +q\log \cos \theta$ is maximum or minimum.

$z=p\log \sin \theta +q\log \cos \theta$
$\frac{dz}{d\theta }=p.\frac{1}{\sin \theta }.\cos \theta +\frac{q}{\cos \theta }(-\sin \theta )$

$\frac{dz}{d\theta }=p.\cot \theta -q\tan \theta$

For maximum or minimum,

$\frac{dz}{d\theta }=0$

$p\cot \theta -q\tan \theta =0$

$\Rightarrow p-q{\tan }^2\theta =0$
$\therefore {\tan }^2\theta =\frac{p}{q}$

or $\theta ={\tan }^{-1}\sqrt{\frac{p}{q}}$

Now, $\frac{d^{2}z}{d{\theta }^2}=-pcose{c}^2\theta -q{\sec }^2\theta$

$\frac{d^{2}z}{d{\theta }^2}=-\left[p\right(1+{\cot }^2\theta )+q(1+{\tan }^2\theta \left)\right]$

${\left(\frac{d^{2}z}{d{\theta }^2}\right)}_{(\theta ={\tan }^{-1}\sqrt{\frac{p}{q}})}=-p(1+\frac{q}{p})-q(1+\frac{p}{q})$
$=-p-q-q-p=-2(p+q).$

which is clearly -ive when ${\tan }^2\theta =\frac{p}{q}$ Hence $z$ is maximum at $\theta ={\tan }^{-1}\sqrt{p/q}$

Hence $y$ is maximum at $\theta ={\tan }^{-1}\sqrt{p/q}$