discuss the extremum of `f(theta)=sin^pthetacos^qtheta, p , q gt0,0ltthetaltpi/2`

Answer

Point of maxima : x = `tan^(-1)sqrt(p/q)`

Explanation

`f(theta)=sin^(p)theta cos^(q) theta,p,q gt 0,0 theta lt theta lt pi//2`
`f'(theta)=p sin ^(p-1) theta cos theta cos^(q-1) theta sin theta sin^(p) theta`
`=sin^(p-1) theta cos^(q-1)theta p cos^(2)theta-q sin^(2 theta)`
Let `f'(theta)=0`
i.e `sin theta =0 or cos theta =0` (not possible)
or p `cos ^(2)theta-p sin^(2)theta =0 or tan^(2) theta = p//q or tan theta = sqrt(p)/(q)`
`(tan theta ne -(sqrt(p))/(q) as 0 lt theta lt pi//2)`
check for extremum
When the only point of extremum is point of maxima
Thus `theta =tan^(-1)sqrt(p)/(q)` is point of maxima when
`tan theta =sqrt(p)/(q) ,cos theta =(sqrt(q))/(sqrt(p+q))` and sin `theta =(sqrt(p))/(sqrt(p+q))`
Hence maximum value `f_(max) =(sqrt(q))/sqrt(p+q)^(q)sqrt(p)/sqrt(p+q)^(p)`
`=p^(q^(q))/(p^(p+q))^(1//2)`