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Sometimes, the constraint is not a compact set. As a result, the local minimum may not be global.

For example, $ f=x^2+y^3$ subject to constraint $ x+y=4/3$. Using Lagrange multiplier method, I calculated local minimum at $(x,y)=(\frac23,\frac23)$. But I don't know what to do next.

I cheated a bit and looked at Wolfram Alpha's plot that shows that there is no global minimum subject to the constraint.

http://www.wolframalpha.com/input/?i=min+x^2+y^3+such+that+{x+y=4/3}

Is there any way to get this result (no global minima) without graphing?

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Choose $y$ small, e.g. $y = -100$, then choose it even smaller $y = -100000$.. –  Lord Soth Jun 28 '13 at 18:25
    
The line $x + y = \frac43$ is not parallel to the $x$-axis, so you get arbitrarily small (and large) $y$ values. Since $x$ is asymptotically $-y$ on that line, the $x^2$ term cannot compensate the $y^3$ term. –  Daniel Fischer Jun 28 '13 at 18:25
    
Thanks Lord and Daniel! –  Amoksh Jun 28 '13 at 18:39

2 Answers 2

up vote 1 down vote accepted

Sure. Find points where the values of $f$ are smaller than the values of $f$ at all the critical points. (For some functions, it's not so clear from a computer graph, unless you know what you're looking for.)

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Isn't that obvious? What's not obvious is how to find those points? Just picking at random? –  Amoksh Jun 28 '13 at 18:30
    
Well, thinking about the qualitative/limiting behavior of the function is the best I can say in general. The $y^3$ is the crucial item here, as you can let $y\to -\infty$ and see what happens. –  Ted Shifrin Jun 28 '13 at 18:35
    
I see, thank you very much. –  Amoksh Jun 28 '13 at 18:38

Simply substituting $x=\frac43-y$ gives you a cubic function $\mathbb R\to\mathbb R$, $y\mapsto y^3+(\frac43-y)^2$. This is has no global extrema as it is onto.

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Thank you, but what if I have more complicated constraint? –  Amoksh Jun 28 '13 at 18:31

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