Finding Absolute Minimum and Absolute Maximum of $f(x,y)=xy$ 
Let $\ f(x,y)=xy$. Use the method of Lagrange multipliers to find the maximum and minimum values of the function f on the circle $\ x^2+y^2=1$

First we note that the function $f$ is continuous and the set $S={(x,y):x^2+y^2=1}$ is compact, hence extrema are guaranteed. 
Using the method Lagrange multipliers, I set $\nabla f=\lambda\nabla g$, where $g(x,y)=x^2+y^2-1$. Following through the calculations, I arrived at four critical points:
$$\Big(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big),\Big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big),\Big(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\Big),\Big(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\Big)$$
Substituting these points into the function $f$, I obtained a maximum at 
$$\Big(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{\sqrt{2}}\Big)=\frac{1}{2}$$
and a minimum at
$$\Big(\pm\frac{1}{\sqrt{2}},\mp\frac{1}{\sqrt{2}}\Big)=-\frac{1}{2}$$

My question is, how do we now find the absolute maximum and absolute minimum of the function $f$ on the unit disc $x^2+y^2\leq 1$?

My attempt so far:
We want to find all the critical points. So to find stationary points, we set 
$$\nabla f=\vec{0}$$
Solving this, we find that $(0,0)$ is a stationary point.
So, $f(0,0)=0$.
Hence the absolute maximum is $\frac{1}{2}$ and the absolute minimum is $-\frac{1}{2}$, as these are all the critical points of $f$.
This does not sit well with me, as I am unsure of my working/logic. Can this be improved on?
 A: You have the minimization optimization problem of the function $f(x,y) = xy$ over the space $S= \{ (x,y) \in \mathbb R^2 : x^2 + y^2 = 1\}$. A known way to deal with Lagrange multipliers is by the Kuhn-Tucker Lagrange method. 
First of all, observe that $f(x,y)$ is continuous and smooth and that the space $S$ is compact. Thus, this means that there exists  a minimum $(\bar{x},\bar{y})$ for $f(x,y)$ in $S$.
By the Kuhn-Tucker Lagrange method, we yield :
$$f_0(x,y) = xy, \; \; f_1(x,y)= x^2+y^2-1$$
and then the K.T.L. system :
$$\begin{cases} \nabla f_0 + \lambda_1\nabla f_1 = 0 \\ \lambda_1 f_1 =0 \end{cases} \Rightarrow \begin{cases} \begin{bmatrix} y \\ x \end{bmatrix} + \lambda_1\begin{bmatrix} 2x \\ 2y \end{bmatrix} =0 \\ \lambda_1(x^2 + y^2 -1) \;= 0\end{cases} $$
Check cases for $\lambda_1 = 0$ and $\lambda_1 >0$ and then you'll yield the same results. (Maximum is given for applying the same method for $-f(x,y)$ or simply you yield the same points as you did.
Now, if there existed another minimum or maximum, it should satisfy the K.T.L. problem. Since no other point satisfies it, these are all the minimums and maximums. Observing that you have two possible minimum and maximum points (since the values are equal) for $f(x,y)$ over $S$, this means that you have a total maximum and minimum at both of the points each time.
A: The method of Lagrange multipliers tells us the the local extremes of the restriction of $f$ to circle are the ones that you got. So, the absolute maximum and the absolute minimum must be attained at some of them (the absolute maximum and the absolute minimum are local extremes). Since $f$ takes the value $\frac12$ in two of them and no value greater than $\frac12$, $\frac12$ is necessarily the maximum. The same argument applies to the minimum.
A: The Lagrange method gives you the local extrema of the function constrained to the boundary *. The stationary points give you the local extrema, and you consider those inside the boundary.
Then the global extrema are achieved by the local extrema that yield the largest/smallest values.

*Alternatively, you could use a parametric equation of the boundary, let $x=\cos t,y=\sin t$, and find the local extrema of $\cos t\sin t$, which occur at $t=\dfrac{k\pi}4$, giving values $\pm\dfrac12$.
