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Suppose we are looking at the following:

$$ \text{minimize} \ f(x) = x^2+y^2 \\ \text{subject to} \ \ x+y-2 \geq 0$$

Would there only be one Lagrange multiplier corresponding to the single constraint? Or would there be two Lagrange multipliers corresponding to $x$ and $y$?

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There would only be one. By the way, note that $x^2+y^2=(1/2)[(x+y)^2+(x-y)^2]\ge (1/2)(x+y)^2$, with equality when $x=y$. So the minimum is reached when $x=y=1$. –  André Nicolas May 9 '12 at 23:24
    
In general, each constraint gives you one Lagrange multiplier, no matter how many variables there are. –  Robert Israel May 9 '12 at 23:27

1 Answer 1

The number of Lagrange multipliers only depends on the number of constraints you have and not on the number of unknowns. For this problem, you can get the solution graphically as shown below.enter image description here

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The plot was generated in grapher. –  user17762 May 9 '12 at 23:39
    
The graph adds a great deal to the insight the OP will obtain from the answer. –  André Nicolas May 10 '12 at 1:24

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