Question

$\text{Show that} \: sin^{p}\theta\cos^{q}\theta\;\;\text{is maximum, if} \\ \: \: \theta=\tan^{-1}\sqrt{\frac{p}{q}}$

Answer 1

$\huge\underline\mathfrak\blue{ANSWER}$

$y = {sin}^{p} \alpha . {cos}^{q} \alpha \\ \\ \frac{dy}{d \alpha } = - {sin}^{p} \alpha .(q). {cos}^{q - 1} . \sin \alpha + p. {sin}^{p - 1} . {cos}^{ q } . \cos \alpha \\ \\ arrange \: the \: powers \: of \: sin \: and \\ cos \alpha \\ \\ \frac{dy}{d \alpha } = - q. {sin}^{p + 1} \alpha . {cos}^{q - 1} \alpha + p. {sin}^{p - 1} \alpha . {cos}^{q + 1} \alpha \\ \\ for \: maximum \: value \: \frac{dy}{dx} = 0 \\ \\ q. {sin}^{p + 1} \alpha . {cos}^{q - 1 } \alpha = p. {sin}^{p - 1} \alpha . {cos}^{q + 1} \alpha \\ \\ \frac{ {sin}^{p + 1} \alpha . {cos}^{q - 1} \alpha }{ {sin}^{p - 1} \alpha . {cos}^{q + 1} \alpha } = \frac{p}{q} \\ \\ \frac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } = \frac{p}{q} \\ \\ {tan}^{2} \alpha = \frac{p}{q} \\ \\ \tan \alpha = \sqrt{ \frac{p}{q} } \\ \\ \alpha = {tan}^{ - 1} \sqrt{ \frac{p}{q} }$

• LET THE GIVEN EQUATION EQUAL TO y
• DIFFERENTIATE y WITH RESPECT TO GIVEN ANGLE. APPLY PRODUCT RULE
• PUT THE GIVEN DERIVATIVE EQUAL. TO 0 TO FIND IT'S MAXIMUM VALUE AT PARTICULAR ANGLE.

ANSWER BY....

PURPLE WORLD

#BTSARMY

Answer 2

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