Given the function $f(x,y)=5y\sin(3x)$ find the stationary points.
I found $f_x=15y\cos(3x)$. Solving $f_x=0$, I got $y=0,x=\frac{(2n+1)\pi}{6}$
Similarly, $f_y=5\sin(3x)$. Solving $f_y=0$, I got $x=\frac{n\pi}{3}$
I thought my stationary points would be $(\frac{(2n+1)\pi}{6},0$) and $(\frac{n\pi}{3},0)$. But the solutions say that it is only $(\frac{n\pi}{3},0)$.
Can someone please explain to me what I'm missing hear or not understanding?
On
What you have found is when the derivative in a single direction is 0 you need to find a single point when the derivative of both directions are 0 you want to find when del of your function is equal to 0. del of $f = f_x +f_y$ when this is zero you have a stationary point. https://en.wikipedia.org/wiki/Del
At the stationary point, both derivatives must be equal to zero. If $x=\frac{(2n+1)\pi}{6},\; y=0$, then $$ f_y= 5\sin 3x= 5\sin \frac{(2n+1)\pi}{2}=5\sin\left(n\pi+\frac{\pi}2\right)=\pm 5\ne 0. $$