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Using the Lagrange multipliers method I have to find the absolute maximum and minimum value of $f(x, y)=x^2+y^2-x-y+1$ in the unit disc.

So, I have to find the extremas of $f(x, y)=x^2+y^2-x-y+1$ subject to $x^2+y^2 \leq 1$, or not??

Do we not apply Lagrange multipliers method when we have a function $f(x,y)$ and a constaint $g(x, y)=0$??

So, shouldn't we have to have an equality at the constraint??

But in this case we have an inequality... What do we do??

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It is a closed region, so max and min must occur. They can only occur on the boundary or at critical points of the function. So you can use the following steps:

Step 1: Find all the critical points of the function, and check whether they are in the constraint region.

Step 2: Use regular Lagrange multiplier method on the boundary of the disk.

Then combine the results from the two steps to find the max and min.

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  • $\begingroup$ What do you mean with "closed" region?? That it is bounded?? $$$$ Can it not be that the critical points of the function are on the boundary?? $\endgroup$
    – Mary Star
    May 21 '15 at 17:47
  • $\begingroup$ It is closed in the sense that the region includes its boundary. A continuous function achieves its max and min in a closed region. So there's no case like the function approaches infinity toward the boundary and never achieves its max. If the critical points are on the boundary, it's even easier. You just need to consider the boundary with the regular Lagrange multiplier method. $\endgroup$
    – KittyL
    May 21 '15 at 17:56
  • $\begingroup$ I see... To find the critical points we do the following: $$\frac{\partial{f}}{\partial{x}}=0 \Rightarrow 2x-1=0 \Rightarrow x=\frac{1}{2} \ \ , \ \ \frac{\partial{f}}{\partial{y}}=0 \Rightarrow 2y-1=0 \Rightarrow y=\frac{1}{2}$$ The only critical poin is $\left (\frac{1}{2}, \frac{1}{2}\right )$. We have to check if it satisfies the constraint. $$\left (\frac{1}{2}\right )^2+\left (\frac{1}{2}\right )^2=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \leq 1$$ So this point is a critical point that satisfies the constraint. $\endgroup$
    – Mary Star
    May 21 '15 at 18:14
  • $\begingroup$ We use Lagrange multiplier method for $f(x, y)=x^2+y^2-x-y+1$ subject to $g(x, y)=x^2+y^2-1=0$. $$\nabla f(x, y)=\lambda g(x, y) \Rightarrow (2x-1, 2y-1)=\lambda (2x, 2y) \Rightarrow x=y$$ $$x^2+y^2=1 \Rightarrow x^2+x^2=1 \Rightarrow 2x^2=1 \Rightarrow x^2=\frac{1}{2} \Rightarrow x=\pm \sqrt{\frac{1}{2}}$$ So, the critical points are $\left (\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{2}}\right )$ and $\left (-\sqrt{\frac{1}{2}}, -\sqrt{\frac{1}{2}}\right )$. Is this correct so far?? $\endgroup$
    – Mary Star
    May 21 '15 at 18:14
  • $\begingroup$ @MaryStar: Yes, it is correct. $\endgroup$
    – KittyL
    May 21 '15 at 18:56

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