# Lagrange multiplier constrain critical point

When using Lagrange multipliers in an inequelity, $$f(x,y) = x^2+y$$ with the constraint $$x^2+y^2 \leq 1.$$ I have to find the critical points inside the "disk" right?

I've done $$f_x = 2x \Rightarrow 2x = 0 \Rightarrow x = 0, \\f_y = 1.$$ So there is no critical point and I should treat this as an equality. Am I right?

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Yes, the extrema should be obtained on $x^2+y^2=1$. – Shuhao Cao Jun 9 '13 at 4:02

The function $f$ is continuous on the disk $D=\{(x,y)\in\mathbb R^2| x^2+y^2\leq 1\}$, which is a compact subset of $\mathbb R^2$. By the theorem of Weiserstrass, $f$ has absolute maximum and minimum on $D$. So you begin searching for stationary points, i.e. solutions of $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}=0$ inside the disk, i.e. on $D^{°}$. You have proven that there is no stationary point. Now you have to consider the restriction $f|_{\partial D}$ of $f$ to the boundary of $D$, i.e. the circle of radius $1$ denoted by $\partial D$.

To do so we use

$x^2+y^2=1 \Rightarrow x^2=1-y^2$,

and $f|_{\partial D}(y)=1-y^2+y$, with $-1\leq y \leq 1$. Then $y=\frac{1}{2}$ is the max for $f|_{\partial D}$, while $y=-1$ is the minimum with $f|_{\partial D}(-1)=-1$. The solutions to our problem are the maxima $(\sqrt{1-\frac{1}{4}},\frac{1}{2})$ and $(-\sqrt{1-\frac{1}{4}},\frac{1}{2})$, with

$f(\sqrt{1-\frac{1}{4}},\frac{1}{2})=f(\sqrt{1-\frac{1}{4}},\frac{1}{2})=\frac{5}{4}$

and the minimum $f(0,-1)=-1$ (this last point can be also found looking at the very definition of $f=x^2+y$; $x^2$ is a non negative quantity and $-1\leq y \leq 1$).

You can check that the above points are maxima/minumum also by definition, i.e. proving that there exist open neighborhoods $U_i$ of the maxima/minumum s.t. for all $(x_i,y_i)$ in $U_i\cap D$ the following inequalities hold

$f(x_1,y_1)-f(\sqrt{1-\frac{1}{4}},\frac{1}{2}))<0$,

$f(x_2,y_2)-f(-\sqrt{1-\frac{1}{4}},\frac{1}{2}))<0$,

and

$f(x_3,y_3)-f(0,1)>0$.

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