# A strange optimization problem

Problem: Determine whether or not the following optimization problem has a solution: $f(x,y,z) = x^{sin(10)} + e^{x+y} + x^3y^{201} + sin(20z)$, subject to: $h(x,y,z) = x^6 - (1/3)x^3y + y^2 + z^4 \le 2011$.

Hint: make use of $|mn| \le a^{-1}|m|^a + b^{-1}|n|^b$, for $a^{-1} + b^{-1} = 1$.

Attempt at solution: There seem to be some random numbers in the setup of the problem which led me to believe that there should be some simple solution to it, however I am unable to find one. By inspection the problem doesn't seem to exclude the possibility of having an answer, since $f$ is continuous and $h^{-1}$ seems to be continuous (I'm not sure about this) so $f$ should take on a minimum on the domain... however this approach makes no use of the hint provided which makes me suspicious if $h^{-1}$ is actually continuous. The other approach I had was using Kuhn-Tucker, but it led nowhere since the gradient of $f$ and $h$ are very messy. Can someone suggest a better way to approach this problem? Or am I missing something obvious? Thanks.

• Forgive me for my lack of knowledge, but how can an optimization problem like this not have a solution? The domain is easily seen to be finite in size (since the most dominant powers of $x, y, z$ all are even with positive coefficient in $h$), and in this domain $f$ has no infinite limits, therefore there must be some subdomain with minimal value (or maximal, depending on what you're trying to solve). – orlp Feb 15 '18 at 15:12
• @orip If the domain is indeed bounded (it's obviously closed) then the minimization problem does have a solution by the extreme value theorem, as I mentioned above. I just find it confusing why the hint was provided and if someone could point out the intended approach to this problem. – Longti Feb 15 '18 at 15:21
• I'm not so sure of the hint, but the function smells of herring of the red variant. – orlp Feb 15 '18 at 15:24

I guess the hint was provided, so you can manually prove that $\{(x,y,z)\in\mathbb{R}^3:\ h(x,y,z)\leq 2011\}$ is bounded. You probably are asked to do the following: $$\frac{1}{3}x^3y\leq \frac{1}{3}(\frac{1}{2}x^6+\frac{1}{2}y^2)$$ Hence $$h(x,y,z)\geq x^6-\frac{1}{6}x^6+ y^2-\frac{1}{6}y^2+z^4 = \frac{5}{6}x^6+\frac{5}{6}y^2+ z^4.$$ This in turn means if $h(x,y,z)\leq C=2011$ you find another constant $C'=C'(C)$, which just depends on $C$ so that $|(x,y,z)|\leq C'$