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?
Best Answer
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.