I am confuse about maximization of a smooth, well behaved objective function $f(x,y, z)$ subject to the constraint that:

$0\leq x \leq 1$

$y+z \leq 200$

$y,z$ are positive.

The function $f(x,y,z)$ is a rational function (second order polynomial divided by second order polynomial)

For this kind of bounded problems, I am wondering if I simply take their partial derivative wrt to x,y,z and set them all to zero. If I am able to solve these 3 equations, am I gaurantee I have arrived a global optima?

$\frac{\partial f(x,y,z)}{\partial x} = 0$ $\frac{\partial f(x,y,z)}{\partial y} = 0$ $\frac{\partial f(x,y,z)}{\partial z} = 0$

And then I also check the boundry condition. Can I gaurantee that I have found the global optima??

I am confused because I am confused about the convex optimization. I didn't understand it. But if the partial derviatives method works, does it mean convex optimization should also work?? Confuse!

  • $\begingroup$ These are very basic questions, this is probably not the best forum to address these questions. The unconstrained problem may have no optimum. Even it you can solve the equations, the resulting points may not be extremisers of the constrained problem. Even if you find a critical point that satisfies the constraints, it may be a $\min$, inflection, etc. I have no idea what you are trying ask about convex problems. $\endgroup$ – copper.hat Jan 5 '18 at 6:06
  • $\begingroup$ If your function is concave and the point whose partial derivatives are equal to zero satisfies your boundary conditions, then you have found a solution to your problem. If not, further work is needed, e.g., Lagrange multipliers and KKT conditions. $\endgroup$ – max_zorn Jan 5 '18 at 6:08
  • $\begingroup$ @copper.hat, My problem is constraint in all variables that I am trying to optimize. My confusion is that I learnt the partial derivatives method in second year, but then somebody told me about the convex optimization, which is graduate level courses. Then I got confused. $\endgroup$ – kou Jan 5 '18 at 6:14
  • $\begingroup$ Other than situations such as Max has mentioned, just finding critical points will not suffice in general. In general, without extra information (such as convexity/concavity, as appropriate), there is no way to determine if a critical point has an extreme (constrained) value. I think your question is too general to get a helpful answer. $\endgroup$ – copper.hat Jan 5 '18 at 6:21
  • $\begingroup$ If we have determine all the critical points, can we just simply compare each critical points to find the global optimum? $\endgroup$ – kou Jan 5 '18 at 6:24

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