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!

Re: Extrema function 2 variables

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

Re: Extrema function 2 variables

Quote:

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.

Re: Extrema function 2 variables

I agree with your first partials:

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

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

However, from these I get the critical points:

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

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.

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

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

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

And so:

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

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

Hence, none of the critical points are extrema.