Find absolute extrema of $f(x,y) = 2\sin(x) + 5 \cos(y)$ in the rectangle with vertices $(0,0), \; (3,0), \; (3,-3), \; (0,-3)$
I am stuck on this problem.
First of all for critical points
$f_x = 2 \cos(x) = 0 \Rightarrow x \in \{ \frac{(2n+1)\pi}{2} \; |\; n \in Z\}$
$f_y = -5\sin(y) = 0 \Rightarrow y \in \{ n \pi \; | \; n \in Z \}$
Now these are infinite many points. Also I am working in cartesian plane but getting coordinates in form of $\pi$ So knowing that a point lies in my region will be too length.
Is there any other method that I can make to solve this?
You're looking to optimize the function with $0 \le x \le 3$ and $-3 \le y \le 0$.
The first step is to locate the zeros of the derivatives which are $x = \left(n + \frac12\right)\pi, y = m\pi$. While there are infinitely many of these, you only need to consider the ones inside your region. Given the boundaries, there is only one, which is $(\pi/2, 0)$
The next step is to optimize the function along the boundaries. There are 4 here: $x = 0, x = 3, y = 0, y = -3$. Along these curves, the function will be dependent only on $x$ or $y$. You'll need to apply the same techniques to optimize these functions.
Conveniently, it just so happens that the critical points of the boundary functions have the same $x$ or $y$ coordinates. $$ f'(x,0) = f'(x,-3) = 2\cos x = f_y $$ $$ f'(0,y) = f'(3,y) = -5\sin y = f_x $$
So you get 3 more critical points $(\pi/2,-3), \ (0,0), \ (3,0)$
Lastly, you'll also have to consider the points at the vertices of the rectangle. In total, this makes for 6 critical points.
The last step is to compute the function value at each of these points. Pick the maximum and minimum out of those, and you're done.