I was reading about partial second derivative test for optimization problems and I came across the example here. I saw the equations have yielded four critical points, but I wasn't able to find those points by myself. First, could you please tell me what is the general approach for finding critical points of multivariable functions? Second, is it possible to find the critical points by having a linear algebraic perspective? If so, do we have to solve for the null space of the points?

Below are the equations to be solved: $$z = f(x, y) = (x + y)(xy + xy^2)$$ $$\frac{dz}{dx} = y(2x + y)(y + 1) = 0$$ $$\frac{dz}{dy} = x(3y^2 + 2y(x + 1) + 1) = 0$$

A side question: I already think the above equations cannot be solved directly using linear algebra. I see linear algebra(Eigenvectors, Eigenvalues, SVD and ...) being used A LOT in convex optimization. How do people actually use linear algebra in non-linear optimization problems? Can you reference me to some resources about this?

Thank you


Let's look at the first equation: for $$\frac{\partial z}{ \partial x} = y (2x + y) (y + 1) = 0$$

either $y = 0$ or $2x + y = 0$ or $y = -1$. Now let's look at the second equation in this light.

$$\frac{\partial z}{\partial y} = x ( 3y^2 + 2y(x+1) + 1) = 0$$.

If $y = 0$ then we have $x(1) = 0$, so $x = 0$. So one critical point is $(0,0)$. If instead $y = -2x$ we have $x(12x^2 - 4x^2 - 4x + 1) = 0$, which you can solve using the quadratic equation (after accounting for the solution $x = 0$. If instead $y = -1$ you can do a similar analysis.

In full generality it might be hard to determine the critical points like this: non-linear systems of equations can be hard. But in this case the first equation was already written as a product of linear factors, so my first step was to analyze each factor.

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  • $\begingroup$ You're right. If finding critical points was easy, there wouldn't be tremendous amount of literature on optimization! So is the only general way of solving equations the way we usually do it in calculus classes? I already know problems having more than 100 variables being solved. In such problems, we will be having a 100+ degree equations! $\endgroup$ – Amir Aug 6 '15 at 22:59
  • $\begingroup$ I did not have enough reputation to upvote your post. I just did it now :) $\endgroup$ – Amir Jan 17 '16 at 0:19

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