Constrained minimization of multi-variable trigonometric function

Question

Let f(x,y) = $\sin^2(x)$+$\arctan^3(xy)$. For every positive integer $n$ and every $M \ge 0$, find $$ \min_{(x,y)\in S_{n,M}} f(x,y), $$

where $S_{n,M} =[0,n \pi]×[−M,M]$,and determine where this minimum is achieved.

I tried to solve this problem by using the first order conditions where I get the derivatives to be complicated trigonometric functions and the solution to either be x does not equal 0 when y = 0 or y=0.

However, I am not convinced that it is the correct way to approach this problem. Any suggestions?

Answer 1

We can start by ignoring the restrictions and computing the stationary points.

$$ 2 \sin x \cos x +\dfrac{3y \arctan^2(xy)}{1+(xy)^2}= 0, \quad \dfrac{3x \arctan^2(xy)}{1+(xy)^2}= 0 $$

looking at the second equation and substituting the candidate values in the first one, we see that the possible solutions are $x = 0$ (for any $y$) and $y=0$ (with $x = \frac{k \pi}{2}, k \in \mathbb{Z}$). So, the stationary points that are interior to $S_{n,M}$ are the points $ (0, \frac{k\pi}{2}), k = 1, \cdots, 2n-1$.

On those interior stationary points the value of $f$ is 1. But the minimum can also be attained at the boundary of $S_{n,M}$.

Case 1: If $x=0, y \in [-M,M]$ then $f=0$

Case 2: If $x=n \pi, y \in [-M,M]$ then $f = \arctan^3(n \pi y)$, that has a minimum for $y=-M$.

Case 3: If $y=M, x \in [0, n\pi]$, $f=\sin^2 x + arctan^3(Mx)$ the attained minimum value is zero.

Case 4:If $y= -M, x \in [0, n\pi]$, $f=\sin^2 x + arctan^3(-M x)$. In this case you have several local minimum points, but the global minimum corresponds to the one found in case 2.

The final conclusion seems to be that the global minimum in $S_{n,M}$ occurs at $(n\pi, -M)$.

Answer 2

One suggestion is to analyse the terms independently: $\sin^2x$ and $\arctan^3(xy)$.

Since $\sin^2x \ge 0$ $\forall x \in [0, n\pi]$, the minimum value that it can achieve is $0$ for $x = n\pi$.

Also, assuming the usual range for principal value, $\arctan^3(xy) \in (-(\frac{\pi}{2})^{3}, (\frac{\pi}{2})^{3})$. Moreover, the function $\arctan^3(z)$ is monotonically increasing for $z \in (-\infty, +\infty)$, so its minimum value will be achieved for $\min \mathscr{D}$ where $\mathscr{D}$ is the domain of the function. In this case, the product of $x$ and $y$ will be minimum for the maximum value of $x$ as $x \in [0, n\pi]$ and the minimum value of $y \in [-M, M]$.

Hence, we can conclude that the minimum value of the required function, with the given range for $x, y$ will be at $x = n\pi$ and $y = -M$ and the minimum value will be $arctan^3(-nM\pi)$.