Lagrange multipliers and the extrema of $f(x,y) = -x^2-y^2+x+y$ Let $f(x,y) = -x^2-y^2+x+y$. Find the extrema of $f$ on unit disk. I have found these. 
But I was asked an extra question: at the Lagrange point $({1\over\sqrt2}, {1\over\sqrt2})$, show that the restriction of $f$ to unit circle $f$ has a local maximum at this point but not a maximum on the whole disk? Any help or hints? 
 A: Your function is 
$$
f(x,y)=-x^2-y^2+x+y
$$
and you would like to maximize this function subject to the points being on the unit disk, i.e.
$$
x^2+y^2\leq 1.
$$
You would need to apply the Kuhn-Tucker optimization, which is a bit different than the lagrange method because you now have an inequality constraint rather than an equality constraint. You would like to take FOCs of the Lagrangean along with the additional complementary slackness condition, i.e.
$$
\mathcal{L}(x,y,\lambda)=f(x,y)+\lambda(1-x^2-y^2)=-x^2-y^2+x+y-\lambda(x^2+y^2-1)
$$
with the optimality conditions being 
$$
\frac{\partial \mathcal{L}}{\partial x}=2(-\lambda-1)x+1=0
$$
$$
\frac{\partial \mathcal{L}}{\partial y}=2(-\lambda-1)y+1=0
$$
and the complementary slackness condition
$$
\lambda(1-x^2-y^2)=0
$$
From the first conditions, we have $x=y$, call it $a$. Then, if  $(1-x^2-y^2)>0$, we would need to have $\lambda=0$, implying that $x=y=1/2$, giving us a critical point. 
If $\lambda>0$, then we would need to have $1-x^2-y^2=0$, which would mean that $x=y=1/ \sqrt{2}$
Notice that, the first critical point would give a function value
$$
f(1/2,1/2)=-1/4-1/4+1/2+1/2=1/2
$$
The second critical point on the other hand would give
$$
f(1/\sqrt{2},1/\sqrt{2})=-1/2-1/2+1/\sqrt{2}+1/\sqrt{2}=\sqrt{2}-1
$$
Notice that $f(1/2,1/2)>f(1/\sqrt{2},1/\sqrt{2})$, and as a result, $f$ is maximized at $(1/2,1/2)$ on the unit disk. The second point is the one where $f$ is maximized on the unit circle and clearly it is not the maximizer on the unit disk.
