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  4. sinp θ cosq θ attains a maximum, when θ =

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Q. $sin^{p}\,\theta \,cos^{q}\,\theta$ attains a maximum, when $\theta = $

Application of Derivatives

Solution:

Let $y= sin^{p}\,\theta \,cos^{q}\,\theta$. Then,
$\frac{dy}{d\theta} = p\,sin^{p-1}\,\theta \, cos^{q}\,\theta +sin^{p}\,\theta \,q\, cos^{q-1}\,\theta \left(-sin\,\theta\right)$
$\Rightarrow \frac{dy}{d\theta } = p\,sin^{p-1}\,\theta cos^{q+1}\,\theta - q\,sin^{p+1}\,\theta \, cos^{q-1}\,\theta$
$\Rightarrow \frac{dy}{d\theta } = sin^{p-1}\,\theta \, cos^{q-1}\,\theta \left(p\,cos^{2}\,\theta - q\,sin^{2}\,\theta \right)$
$\Rightarrow \frac{dy}{d\theta } = sin^{p}\,\theta \,cos^{q}\,\theta \left(\frac{p\,cos^{2}\,\theta - q\,sin^{2}\,\theta }{sin\,\theta \,cos\,\theta }\right)$
$\Rightarrow \frac{dy}{d\theta } = sin^{p}\,\theta \,cos^{q}\,\theta \left(p\,cot\,\theta - q\,tan\,\theta \right)$
For maximum or minimum, we must have
$\frac{dy}{d\theta } =0$
$\Rightarrow sin^{p}\,\theta \,cos^{q}\,\theta \left(p\, cot\,\theta - q\,tan\,\theta \right) = 0$
$\Rightarrow sin^{p}\,\theta = 0$ or
$ \,cos^{q}\,\theta = 0$ or
$ p\,cot\,\theta - q\,tan\,\theta = 0$
$\Rightarrow sin^{p}\,\theta = 0$ or
$ \,cos^{q}\,\theta = 0$ or,
$ tan\,\theta = \sqrt{\frac{p}{q}} = \alpha$
$\Rightarrow \theta = 0$ or,
$ \theta = \frac{\pi }{2}$ or,
$\theta = tan^{-1}\sqrt{\frac{p}{q}} = \alpha \left(say\right)$
Now,
$\frac{dy}{d\theta } = sin^{p}\,\theta \,cos^{q}\,\theta \left(p\,cot\,\theta - q\,tan\,\theta \right)$
$ = y\left(p\, cot\,\theta - q\,tan\,\theta \right)$
$\Rightarrow \frac{d^{2}y}{d\theta^{2} } = \frac{dy}{d\theta } \left(p\,cot\,\theta - q\,tan\,\theta \right)+ y\left(-p\,cosec^{2}\,\theta - q\,sec^{2}\,\theta \right)$
$\Rightarrow \left(\frac{d^{2}y}{d\theta ^{2} }\right)_{\theta = \alpha} = \left(\frac{dy}{d\theta }\right)_{\theta = \alpha } \left(p\sqrt{\frac{q}{p} - q \sqrt{\frac{p}{q}}}\right)$
$+ sin^{p}\,\theta \,cos^{q}\,\theta \left[-p\,cosec^{2}\,\theta - q\,sec^{2}\,\theta \right]$
$\Rightarrow \left(\frac{d^{2}y}{d\theta ^{2} }\right)_{\theta = \alpha } = 0-sin^{p}\,\theta \,cos^{q}\,\theta \left(p\,cosec^{2}\,\theta - q\,sec^{2}\,\theta \right)$
Hence, $y$ is maximum when $\theta = \alpha = tan^{-1} \sqrt{\frac{p}{q}}$.