# Maximizing/minimizing $f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$

Assume that $$f : [0, 2\pi]\rightarrow \mathbb{R}$$ is a function such that $$f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta)$$. Then, how can we maximize/minimize $$f$$?

We can re-parametrize our function $$f$$ by defining another function $$g : [-1, 1]\rightarrow \mathbb{R}$$ function such that for every $$t\in [-1, 1]$$,

$$g(t) = \sqrt{2}\sqrt{1-t^2}-4t$$

$$\frac{dg}{dt} = \frac{d}{dt}\left(\sqrt{2}\sqrt{1-t^2}-4t\right) = \frac{\sqrt{2}t}{\sqrt{1-t^2}} + 4 = 0$$

From which we conclude that $$g$$ attains its maximum/minimum at $$\left(-\frac{2\sqrt{2}}{3}, g\left(-\frac{2\sqrt{2}}{3}\right)\right), (1, g(1))\in \mathbb{R}^2$$ respectively.

• Your approach is fine. But next to the treatment of $0$, $2\pi$ you also have to take into account that we do not have $\cos(\theta)=\sqrt{1-\sin(\theta)^2}$ for all $\theta\in[0,2\pi]$. Jan 17, 2023 at 22:15

Use Cauchy-Schwarz inequality: $$f^2(\theta) = \left(\sqrt{2}\cos\theta - 4\sin\theta\right)^2\le ((\sqrt{2})^2+4^2)(\cos^2\theta+\sin^2\theta)=18\implies -3\sqrt{2} \le f(\theta) \le 3\sqrt{2}$$. Can you conclude the min and max for $$f$$?
$$f(\theta) = \sqrt{2}\cos(\theta)-4\sin(\theta) = \sqrt{18}( \sin\alpha \cos \theta - \cos \alpha \sin \theta)$$ where $$\sin \alpha = \frac{\sqrt{2}}{\sqrt{18}}$$ and $$\cos \alpha = \frac{4}{\sqrt{18}}$$ and hence $$\alpha = arctan \frac{\sqrt{2}}{4}$$. Thus $$f(\theta) = \sqrt{18}\sin(\alpha - \theta) \le \sqrt{18}$$ and the estimate is best possible.
I think, re-parametrazing is waste of time. $$f'(\theta)=-\sqrt2\sin\theta-4\cos\theta=0$$ gives $$\tan\theta=-2\sqrt2$$ which has two solutions $$\frac{\pi} {2}<\theta_1=\pi+\arctan(-2\sqrt2)<\pi$$ and $$\pi<\theta_2=\pi+\theta_1<2\pi$$ in $$[0,2\pi].$$ $$f(\theta_1)=\sqrt2(-\frac13)-4\frac{2\sqrt2}{3}=-3\sqrt2$$ and similarly $$f(\theta_2)=3\sqrt2$$. We need to compare these values with the boundry value(s) $$f(0)=f(\pi)=\sqrt2$$. Overall: $$(\theta_1,-3\sqrt2)$$ is global minumum and $$(\theta_2,3\sqrt2)$$ is global maximum.
I haven't gone through the details of your answer with a fine-toothed comb, but assuming your calculations are correct, you still haven't convinced me that those values are the maximum/minimum values of $$f$$ in the interval $$[0,2\pi]$$.
As well as looking at the stationary points, you need to look at the end points also: that is, $$f(0)$$ and $$f(2\pi)$$, or $$g(-1)$$ and $$g(1)$$, as maximum or minimum might happen at these points rather than the stationary points.
Basically there is a theorem that says that the maximum and minimum of a continuous real function $$\ f: [a,b]\to\mathbb{R}\$$ must occur either at a stationary point or at one of the end points.