Maximum value of $4|\cos x|-3|\sin x|$

Question

How will I find the maximum of $4|\cos x|-3|\sin x|$

The absolute value confuses me, The maximum value would be 5 if it were not there.

Answer 1

$a=|sin x|,b=|\cos x|$ where $a,b\in[0,1]$ we have to maximise $$4a-3b=4a-3\sqrt{1-a^2}=f(a)$$ but $$f'(a)=4+\frac{3a}{\sqrt{1-a^2}}>0$$ hence $$f(a)\le f(1)=4$$

Answer 2

The maximum of your expression cannot exceed $4$, which is obtained when $4|\cos x|$ is maximized and $3|\sin x|$ is minimized independently.

In this case, at $x=n\pi~(n\in\Bbb Z)$, both maximization of the first term and minimization of the second term happen simultaneously. So the maximum value is indeed $4$.

Answer 3

$|\cos (x)| = 1$(max value) for all $x = n\pi, n\in \Bbb Z$
So, $4|\cos (x)| = 4$ is the maximum value possible of the first term.

$3|\sin x| \ge 0$. So, we need the term $3|\sin x|$ to have the minimum value possible as it is being subtracted from the first term and that value is zero. This again occurs at $x = n\pi, n\in \Bbb Z$.

So, $4|\cos x| - 3|\sin x|$ attains a max. value of $4-0 = 4$ at $x = n\pi, n\in \Bbb Z$.