I am studying the field of optimisation and have came across the following question concerning local minimisers and I demonstrate my work below I am confused with if the minimisers I have found are correct.
Question Consider the objective function: $$f(x,y) = 2x^{3} + xy^{2} + 5x^{2} + y^2.$$
Find the points which satisfy the first-order necessary conditions for local extremisers. Which of these points are local minimisers and maximisers, and does f(x,y) have any global minimisers?
My attempt
By computing the partial derivatives and setting the results equal to 0:
$$ \frac{\partial f}{\partial x}(x,y) =\ 6x^{2} + y^{2} + 10x =0 $$
$$ \frac{\partial f}{\partial y}(x,y) =\ 2xy + 2y = 0 $$
The points which I have found are:
x1 = [0,0]
x2=[-5/3 , 0]
x3 = [-1, 2]
x4 = [-1, -2]
To try and find the minimisers I have done:
$$H = \left[\begin{array}{cc}{12x+10} & {2y}\\ {2y} & {2x+2}\end{array}\right]$$ $$
Evaluating the points:
x1, has two eigenvalues which are both 0 - this is therefore a saddle point.
x2, has eigenvalues of -4/3 and -10, these values are real and negative, this is therefore a local maximiser.
x3, has eigenvalues: -1+-sqrt(17), there is one eigenvalue which is negative and one which is positive. This is therefore a saddle point.
x4, has the same eigenvalues as x3 and so I come to the same conclusion.
Therefore I have found that x2 = [-5/3, 0] is a global maximiser and there are no global minimisers.
In finding any global minimisers I consider that I need to show that the local extremisers I need to find the values which are less than or equal to f(x,y) although I don't really know what to do at this stage.
Thank you for any assistance, I am grateful for any help.
-S
I agree with your partial derivatives, your critical points, and your Hessian.
You say that, at $x_1$, $H$ has $0$ as both its eigenvalues (I agree), but this does not imply that $x_1$ is a saddle point. The second derivative test has failed in this case, and further study is required to determine the nature of the critical point. As it turns out, it's a local minimum, but it takes a bit more work to prove. Note that $$\operatorname{det} H = (12x + 10)(2x + 2) - 4y^2.$$ For $(x, y)$ close, but not equal to $0$, note that this determinant is going to be strictly positive (the whole expression tends to $20$ as $(x, y) \to 0$) and $(12x + 10)(2x + 2) > 0$, again for $x$ sufficiently close to $0$. This tells us that $H$ is positive semidefinite on a neighbourhood of $(0, 0)$, which implies that $f$ is convex on this neighbourhood, and so $(0, 0)$ must be a local minimum.
I agree with the other eigenvalue computations and their conclusions.
As for the conclusions about global extrema, I disagree that $(-5/3, 0)$ is a global maximiser. I don't really understand how you did get to this conclusion, but it's not right. If you consider $f(x, y)$ along the line $y = 0$, you get $$f(x, 0) = 2x^3 + 5x^2,$$ which is a cubic function in $x$. As $x \to \infty$, $f(x, 0) \to \infty$ and as $x \to -\infty$, $f(x, 0) \to -\infty$. Therefore, the values in the range of $f$ become as large and positive as you like, and also become as large and negative as you like, proving that none of the extrema are global.