# Find critical and saddle points for a multivariable function

The function is $f(x, y) = x^2 + 2y^2 + 2xy - 4y + 15$

here's what I did,

$$f_x = 2x + 2y$$ $$f_y = 4y + 2x - 4$$ $$f_{xx} = 2$$ $$f_{yy} = 4$$ $$f_{xy} = 2$$

Putting $f_x = 0$ and $f_y = 0$, we find that $x = -2$ and $y = 2$

$$D(x, y) = f_{xx}(x, y)f_{yy}(x, y) - [f_{xy}(x, y)]^2$$ $$D(-2, 2) = 2\cdot{}4-(2)^2 = 4$$

Since $4 > 0$, it is the local minima

How do I determine the saddle point here? and does this mean the function doesn't have a local maxima? Also how to determine if the local minima is also global?

• Small note: since $D = 4 > 0$ and $f_{xx} > 0$, this means the stationary point is a local minimum. – Theo Bendit Jun 23 '17 at 22:15
• You have a paraboliod. What do you know about paraboliods? – Doug M Jun 23 '17 at 22:27

## How do I determine the saddle point here?

There is no saddle point. You found there was exactly one stationary point and determined it to be a local minimum. For there to be a saddle point, you'd need to find another stationary point, and compute $D < 0$.

## And does this mean the function doesn't have a local maximum?

Yes. Same issue: there's no other stationary points, so there cannot be other local maxima or minima.

## Also how to determine if the local minima is also global?

Good question. It's not as easy as determining local extrema. One method would be to selectively factorise the function:

\begin{align*} f(x, y) &= x^2 + 2y^2 + 2xy - 4y + 15 \\ &= x^2 + 2xy + y^2 + y^2 - 4y + 4 + 11 \\ &= (x + y)^2 + (y - 2)^2 + 11, \end{align*}

which is a sum of squares, which is minimised when the squares are $0$ (yielding the minimum you found earlier). But now, we see that the minimum is actually global, rather than local, because these squares can never be less than $0$.