Find local minima, maxima, and saddle points for $f(x,y) = \sin x + \cos y + \cos(x-y)$ when $0\le x\le\frac\pi2$ and $0\le y\le\frac\pi2$

Question

What I have done for this problem:

I differentiated $f(x,y)$ with respect to $x$ and $y$. Then, I set $f_x$ and $f_y$ to $0$ to find stationary points.

$$f(x,y) = \sin(x) + \cos(y) + \cos(x-y)$$ $$f_x = \cos(x) - \sin(x-y)$$ $$f_y = -\sin(y) + \sin(x-y)$$ $$f_x = f_y = 0$$ $$\cos(x) = \sin(x-y)$$ $$\sin(y) = \sin(x-y)$$ $$\cos(x) = \sin(y)$$ Note: $\cos(x) = \sin(y)$ implies $\cos(y) = \sin(x)$. $$\sin(\pi/2 - x) = \sin(y)$$ $$y = \pi/2 - x \,\,\{0\le x \le\pi/2, 0\le y \le\pi/2\}$$

Using the Second Derivative Test, I categorized the points on the line segment ($y=\pi/2 - x$).

$$f_{xx} = -\sin(x) - \cos(x-y)$$ $$f_{yy} = -\cos(y) - \cos(x-y)$$ $$f_{xy} = \cos(x-y)$$ $$f_{yx} = \cos(x-y)$$

$$D = f_{xx}f_{yy} - f_{xy}f_{yx}$$ $$D = \cos(y)(\cos(y) + 2\cos(x-y))$$

Finally, the Second Derivative Test is inconclusive when $y = \pi/2$ and $x = 0$, and for all the other points on the line segment $f(x,y)$ when $0 \lt x \le \pi/2$ and $0\le y \lt \pi/2$ represents local maximum because $D \gt 0$ and $f_{xx} \lt 0$. This does not make sense because when I graph $f(x,y)$ on Desmos 3D graphing calculator it only shows one local maximum and one local minimum (https://www.desmos.com/3d/326ec8bd3d).

Answer 1

The problem seems more difficult than necessary because you haven't found the critical point properly.

It turns out that there is a unique critical point in the domain $[0, \pi/2] \times [0, \pi/2]$ that lies at the intersection of the curves $f_x = 0$ and $f_y = 0$.

Let's solve the first equation: $$ \cos x - \sin(x-y) = 0. $$ Using the cofunction identity $\cos x = \sin\bigl(\frac{\pi}{2} - x\bigr)$, our equation becomes $$ \sin\bigl(\tfrac{\pi}{2} - x\bigr) = \sin(x-y). $$ Since the arguments satisfy $0 \leq \frac{\pi}{2} - x \leq \frac{\pi}{2}$ and $-\frac{\pi}{2} \leq x - y \leq \frac{\pi}{2}$ on our domain, and the sine function is one-to-one there, we can conclude that $$ \tfrac{\pi}{2} - x = x - y $$ or $$ y = 2x - \tfrac{\pi}{2}. $$ All the points on this line satisfy $f_x = 0$.

I will let you solve $f_y = 0$.

We get $f_y = 0$ at points on the line $y = \frac12 x$.

Now, we solve these two linear equations simultaneously to find the critical point

$$(x, y) = \bigl(\tfrac{\pi}{3}, \tfrac{\pi}{6}\bigr).$$

Here's a picture of level curves, lines $f_x = 0$ and $f_y = 0$, and the critical point.

From there, you can use the second derivative test to conclude:

$f$ achieves a maximum value of $\frac{3\sqrt{3}}{2}$ at the critical point.