Maximum value of $x\cos(x)$ on $[0, \pi/2]$

Question

I would like to find the maximum value of : $f(x) = x\cos(x)$ on $[0, \pi/2]$. The huge problem I have is that $x$ is increasing on this domain and $\cos(x)$ is decreasing.

I tried to find the derivative which is : $-\sin(x)x + \cos(x)$ yet finding its sign is as difficult as the original problem.

So is there a shortcut I am missing here ?

Thank you !

Answer 1

As you noted, you must solve

$$ f^\prime(x)=\cos(x)-x\sin(x)=0 $$

on the interval $[0,\pi/2]$.

I suggest expressing $\cos(x)-x\sin(x)=0$ as $1-x\tan(x)=0$ and solving

$$ g(x)=1-x\tan(x)=0 $$

by Newton's method. A sketch of the graph of $f(x) = x\cos(x)$ indicates a solution slightly larger than $\dfrac{\pi}{4}$ so a good choice for an initial guess for the solution would be $x_0=\dfrac{\pi}{4}$

Newton's method gives the recursion \begin{eqnarray} x_{n+1}&=&x_n-\frac{g(x_n)}{g^\prime(x_n)}\\ &=&x_n+\frac{1-x_n\tan(x_n)}{\tan(x_n)+x_n\sec^2(x_n)} \end{eqnarray}

This gives

$$ x_1=\frac{\pi}{4}+\frac{1-\pi/4}{1+\pi/2}\approx0.869 $$

which is already close the the solution which is closer to $0.86$.

One more iteration should give a reasonable approximation.