How would I find the direction of fastest increase of $f(x,y) = \sin(x + y)xy$ starting at $(-\frac{\pi}{2}, \frac{\pi}{2})$?

Question

So far I've got:

Let $v$ be the direction where $v = (x_1, x_2)$

$\nabla f = (\sin(x+y)y + \cos(x+y)xy, \cos(x+y)xy+\sin(x+y)x)$

$\nabla f (-\frac{\pi}{2}, \frac{\pi}{2}) = (-\frac{\pi^2}{4}, -\frac{\pi^2}{4})$

So I expect that the directional derivative starting at $(-\frac{\pi}{2}, \frac{\pi}{2})$ in the direction of $v$ is:

$\nabla f(-\frac{\pi}{2}, \frac{\pi}{2}) \cdot v = -\frac{\pi^2}{4}x_1 - \frac{\pi^2}{4}x_2$

So I guess that I need to maximise $-\frac{\pi^2}{4}x_1 - \frac{\pi^2}{4}x_2$ in order to complete the question but I am not completely sure.

Is this correct? and what would I need to do from there?

Thanks very much in advance.

Answer 1

Just take the gradient $(-\pi^2/4,-\pi^2/4)$, it is the direction of the fastest increase; as a direction, you may take any positive multiple of this vector, such as $(-1,-1)$ or $(-1/\sqrt{2}, -1/\sqrt{2})$.

If you like the complicated way, yes, you can maximize $-(\pi^2/4) x_1 - (\pi^2/4) x_2$ subject to $x_1^2+x_2^2=1$. The solution is of course the normalized gradient $(-1/\sqrt{2}, -1/\sqrt{2})$.