Prove that a point is not local maximum or local minimum of the function $f(x,y)$

maxima-minimamultivariable-calculus

I was faced with this question:

Given that $f(x,y)=x^3-y^3-3x^2+6y^2+3x-12y+8$.
Prove that $(1,2)$ is a critical point of $f$, and that it's not a local minimum or a local maximum.

It's my first time dealing with a question like this, I have found the point and found that $f_{xx}f_{yy} – (f_{xy})^2 = 0$.
And now I need to prove that it's not maximum or minimum, how do I do that? The only way I thought about is just substituting points until I get $f(x,y) > f(1,2)$ and another point $f(x,y) < f(1,2)$.
But that obviously isn't an efficient way and might be wrong.

I would appreciate any help in how to deal with this question and questions like these in general whenever I get a point that isn't a min/max/saddle, how can I decide what point is it?

Best Answer

On the vertical line $x=1$, we have $$ f(x,y)=f(1,y)=-y^3+6y^2-12y+9=-(y-2)^3+1 $$ which is strictly decreasing, hence $f$ does not have a local min or max at $(1,2)$.

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

x^2+y^2(1-x)^3

[Math] Property of critical point when the Hessian is degenerate

Informally, you only need to look at the function when $x$ and $y$ are small. In that case, the 3rd order element $x^2y$ and the 4th order elements $xy^3$ can be neglected in front of the 2nd order element $5x^2$. So $f(x,y) \approx 5x^2$, which looks like a parabolic cylinder in the $Oy$ direction.

More formally, replace $f$ by the first terms of a Taylor series : $$ f(x,y) = f(0,0) + \frac {\partial f} {\partial x} x + \frac {\partial f} {\partial y} y + \frac {\partial^2 f} {\partial x^2} x^2 + \frac {\partial^2 f} {\partial x \partial y} x y + \frac {\partial^2 f} {\partial y^2} y^2 +....$$

keeping only the terms of the smallest degree, which gives you an approximation of the function in the neighborhood of $(0,0)$.

In your case, your $f$ is a polynomial, so the Taylor expansion is itself.

Best Answer

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

[Math] Property of critical point when the Hessian is degenerate

Related Question