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Here is an minimization problem involving $2$ variables, $x$ and $y$. And there are $2$ functions, $g(x,y)$ and $h(x,y)$.

The goal of this minimization problem is to find out the $(x,y)$ which minimizes both $g(x,y)$ and $h(x,y)$ at the same time.

I just know how to solve simple optimization problem subject to equation constraints, like to minimize $e(x,y)$ while subject to $f(x,y) = k$, for this kind of problem, use Lagrange Multiplier to solve it.

But here, the problem isn't like the Lagrange Multiplier type, is it? How to solve it?

avocado
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  • What does it mean to minimize both $g$ and $h$ at the same time? – André Nicolas Jul 15 '13 at 14:12
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    You can't necessarily minimize both $g$ and $h$ simultaneously, and even if you can it still amounts to minimizing $g$, minimizing $h$, and seeing if they have the same minimum. You could loosen the conditions of the problem by trying to minimize $g(x,y)+h(x,y)$, or perhaps $a g(x,y)+ b h(x,y)$ where $a$ and $b$ are weighting constants. – Ben Grossmann Jul 15 '13 at 14:17
  • OP, can you clarify a point of confusion. Are you aware that the method of lagrange multipliers requires only that ${g,h} \in \mathcal{C}^1$? Are you unsure of whether the method is applicable? or do you want an alternative solution to lagrange's method? – franklin Jul 15 '13 at 14:27

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Unless you want to settle for solving two separate problems (which is a nice alternative in some curcumstances), try combining $g,h$ into one function to minimize. In practice you would probably like to minimize $g+h$ or $g \times h$. This way if the same point minimizes both $g$ and $h$, it will also minimize the sum.

The interesting part is if the same point does not minimize both. Then, you have to define more clearly what is means to minimize both -- perhaps you vcan trade off some in $g$ to get a better value in $h$... This tradeoff will be defined when you combine the functions - either by addition or by multiplication. If you want more control, consider minimizing $Cg(x,y) + h(x,h)$ for constant $C$ which gives an explicit price of a 1-step in $h$ as a $C$-step in $g$...

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