Maximum value of a function. I am not able to check double derivative.

Can someone explain me how $\sin^p x \cos^q x$ attains maximum at $\tan^2 x = \frac pq$.

I am not able to check whether double derivative is positive or negative.

Question

Show that $$\sin^p\theta\cos^q\theta$$ attains a maximum when $$\theta=\tan^{-1}\sqrt{(p/q)}$$

Solution

Ley $y=\sin^p\theta\cos^q\theta$. For a maximum or minimum of $y$, we have $\frac{\text dy}{\text dx}=0$

\begin{align}p\sin^{p-1}\theta\cos^{q+1}\theta-q\sin^{p+1}\theta\cos^{q-1}\theta&=0\\ \sin^{p-1}\theta\cos^{q-1}\theta(p\cos^2\theta-q\sin^2\theta)&=0\end{align}

Therefore \begin{align}\sin\theta&=0\\ &\Downarrow\\ \theta&=0\\ \text{or } \cos\theta&=0\\ &\Downarrow\\ \theta&=\frac \pi 2\\ \text{or }\tan^2\theta&=\frac pq\\ &\Downarrow\\ \theta&=\tan^{-1}\sqrt{p/q}\end{align}

Now $y=0$ at $\theta=0$ and also at $\theta=\frac\pi2$

When $0<\theta<\frac\pi2$, $y$ is positive

Also, $\tan^{-1}\sqrt{p/q}$ is the only value of $\theta$ lying between $0$ and $\frac \pi2$ at which $\frac{\text d}{\text dx}=0$.

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

This can be seen from the graph of $y$

Solution 1:

This is a nice application of logarithmic differentiation.

Consider $$f=\sin^p(x) \cos^q(x)\implies \log(f)=p \log(\sin(x))+q \log(\cos(x))$$ Differentiate both sides $$\frac{f'}f=p \cot(x)-q \tan(x)=\frac{p-q \tan^2(x)}{\tan(x)}$$ and then the conditions for an extremum as already given in answers.

But, we need to check the second derivative : start with $$f'=f \left(\frac{f'}f \right)$$ and differentiate using the product rule $$f''=f'\left(\frac{f'}f\right)+f \left(\frac{f'}f \right)'$$ But, at the extremum $f'=0$ which reduces the problem to the sign of $$f \left(\frac{f'}f \right)'=-f\,(p \csc ^2(x)+q \sec ^2(x))$$

Edit

If you want to continue with some fun, using $$\sin(\tan^{-1}(t))=\frac{t}{\sqrt{t^2+1}}\qquad \cos(\tan^{-1}(t))=\frac{1}{\sqrt{t^2+1}}$$ and replacing $t$ by $\sqrt{\frac{p}{q}}$ the maximum value of $f$ is given by $$f_{max}=\sqrt{\frac{p^p \, q^q}{(p+q)^{(p+q)}}}$$

As said in comments, this works for any value of $p>0$, $q>0$, $p$ and $q$ being integers, rational or non rational numbers.

Considering the case where $p+q=k$, using again logarithmic differentiation we should find that $$\frac{f'_{max}}{f_{max}}=\frac 12 \log \left(\frac{p}{k-p}\right)$$ which is positive if $p>\frac k 2$ and negative otherwise.

Solution 2:

Your picture shows a proof that $f'(x)=0$ at $x=x_0=\arctan\sqrt{p/q}$. As $p$, $q\ge1$, $0<x_0<\pi/2$. Think of the graph of $f$ on $(0,\pi/2)$: $f(0)=f(\pi/2)=0$ and $f(x)>0$ if $0<x<\pi/2$. Then $f$ certainly has a maximum in the open interval $(0,\pi/2)$. But the calculation in the picture shows the only turning point is at $x_0$. So $f$ has a maximum at $x_0$. You don't need to consider $f''(x_0)$.

Solution 3:

When you want to maximize $Sin^p x Cos^q x$ you can equally maximize $Sin^{2p} x Cos^{2q} x$ so ,suppose we need to maximize $$f=Sin^{2p} x Cos^{2q} x=\\ (Sin^2 x)^ (Cos^2x)^q $$ we know $$sin^2x+cos^2x=1 $$ so ,put down $cos^2x=1-sin^2x$ in $f$ $$f=(Sin^2 x)^p (Cos^2x)^q=(Sin^2 x)^p (1-in^2x)^q=X^pY^q$$with $$X+Y=1$$ now ,take f' $$f=X^p(1-X)^q \to \\f'=pX^{p-1}(1-X)^{q}-qX^{p}(1-X)^{q-1}=0\\ X^{p-1}(1-X)^{q-1}(p(1-X)-qX)=0\\$$form here $$X=0,1,\frac{p}{p+q}$$ $0,1$ does not make maximum $f(0)=0,f(1)=0$ so $f(\frac{p}{p+q})$ is max, when you return to begining ,you will see $$X=sin^2 x=\frac{p}{p+q}$$ so find $$cos^2x=1-X=1-\frac{p}{p+q}=\frac{q}{p+q}$$ and finally $$tan^2x=\frac{sin^2x}{cos^2x}=\frac{\frac{p}{p+q}}{\frac{q}{p+q}}=\frac{p}{q}$$