Thread: 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!

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

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.

I agree with your first partials:





However, from these I get the critical points:



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.

(note we need the negative sign as .





And so:

thus:



Hence, none of the critical points are extrema.