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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?

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

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