# Thread: Extrema function 2 variables

1. ## Extrema function 2 variables

Hello!

I need to calculate the stationary points and determine their nature of f(x,y) = x + ysin(x).

I'm a little confused by setting δf/δx and δf/δy equal to zero and by that finding the stationary points...

Can someone solve this for me? Thanks!

2. ## Re: Extrema function 2 variables

What do you get when you equate the first partials to zero?

3. ## Re: Extrema function 2 variables

Originally Posted by MarkFL2
What do you get when you equate the first partials to zero?
Here's what I got:

f(x,y) = x + ysin(x)

∂/∂x = 1 + ycos(x)
∂/∂y = sin(x)

So,

1 + ycos(x) = 0 -> y = 1 when even and -1 when odd
sin(x) = 0 -> x = kπ

-> Critical points (kπ,1) and (kπ, -1)

Second derivative test for P(kπ,1) :
∂²/∂x² = ysin(x) -> in P: -1sin(π) = 0
∂²/∂y² = 0 -> in P: 0
∂²/∂x∂y = cos(x) -> in P: cos(kπ) = -1 or 1

So Δ = (∂²/∂x²)(∂²/∂y²) - (∂²/∂x∂y)² = (0)(0) - (±1)² = 0 - 1 = -1
Thus Δ < 0 -> P(kπ,1) is a saddle point.

4. ## Re: Extrema function 2 variables

I agree with your first partials:

$f_x(x,y)=1+y\cos(x)=0$

$f_y(x,y)=\sin(x)=0$

However, from these I get the critical points:

$(2k\pi,-1),\,((2k+1)\pi,1)$

The sign of the y-coordinate is reversed from what you stated, but this doesn't actually affect the conclusion drawn from the second partials test.

$f_{xx}(x,y)=-y\sin(x)$ (note we need the negative sign as $\frac{d}{dx}(\cos(x))=-\sin(x)$.

$f_{yy}(x,y)=0$

$f_{xy}(x,y)=\cos(x)$

And so:

$D(x,y)=-\cos^2(x)$ thus:

$D(2k\pi,-1)=D((2k+1)\pi,1)=-1$

Hence, none of the critical points are extrema.