# Math Help - difficult stationary points question

1. ## difficult stationary points question

the interval is x and y between 0.25pi and 0.5pi (the square thing kind of confused me!)

I got

∂u/∂x = siny sin(2x+y) =0

siny = 0

y = 0 ± npi (not in interval so doesn't count ; npi means integer number of pi)

OR

2x +y =0

Also,

∂u/∂y = sinx sin(2y+x) =0

Again sinx =0 ; x = 0 ±npi is outside the given interval

2y + x =0

if you try to find x:

2x -0.5x = 0 ±npi

x = n * 2/3 * pi

even with n = 1 this is outside the interval!!

2. ## Re: difficult stationary points question

First, you got $2x+y = n_1\pi$. Then you got $2y+x = n_2\pi$. So, solving for $y$ in the first equation gives: $y = n_1\pi - 2x$. Plugging into the second gives:

$2(n_1\pi - 2x)+x = n_2\pi$

$(2n_1 - n_2)\pi = 3x$

Note that the map $\mathbb{Z}^2 \mapsto \mathbb{Z}$ that takes $(n_1,n_2) \mapsto (2n_1-n_2)$ is a surjection, so we can replace $(2n_1-n_2)$ with $n$:

So, $3x = n\pi$ where $n$ is any integer.

$x = \dfrac{n\pi}{3}$, and $\dfrac{\pi}{3} \in \left[\dfrac{\pi}{4},\dfrac{\pi}{2}\right]$

Plugging in: $y = n_1\pi - 2\dfrac{\pi}{3} = \dfrac{(3n_1-2)\pi}{3}$

So, if $y = \dfrac{\pi}{3}$, it too is in the interval. Those are the only values for $x$ and $y$ that work.