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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?

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    $\begingroup$ Could be beneficial to parameterize the unit circle $(x,y)=(\cos t,\sin t)$ and consider optimizing $f(x,y)=xy=\cos t\sin t=\frac12\sin2t$ for $t\in[0,2\pi]$. $\endgroup$ – A.Γ. Jun 13 '18 at 8:10
  • $\begingroup$ an observation is that under the constraint: $(x+y)^2 = x^2+y^2+2xy = 1 + 2xy$ and the solution becomes trivial. $\endgroup$ – Stefan Jun 13 '18 at 8:17
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    $\begingroup$ You probably mean the unit disk $x^2+y^2\le 1$ ? $\endgroup$ – Yves Daoust Jun 13 '18 at 8:18
  • $\begingroup$ Yes, question has been edited. Thanks! $\endgroup$ – user557493 Jun 13 '18 at 8:21
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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.

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  • $\begingroup$ Thank you for your response! Just so that I understand, are there 5 critical points? Two of which $f$ takes the value $\frac{1}{2}$, two of which $f$ takes the value $-\frac{1}{2}$ and a final critical point which takes the $f$ value of $0$? I am unsure of how many critical points there are. $\endgroup$ – user557493 Jun 13 '18 at 8:09
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    $\begingroup$ No. There are only four critical points. None of the points at which $f$ takes the values $0$ (that is, the points $(\pm1,0)$ and $(0,\pm1)$) is critical. $\endgroup$ – José Carlos Santos Jun 13 '18 at 8:11
  • $\begingroup$ The following question asks to classify every stationary point of the function $f$. Are you able to provide some insight into this question? Is $(0,0)$ a stationary point? How many stationary points are there? $\endgroup$ – user557493 Jun 13 '18 at 8:17
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    $\begingroup$ Is $(0,0)$ a saddle point? Also, if $f(0,0)\geq\frac{1}{2}$ would this be the absolute max? I thought the absolute max of $f$ was the largest value of every critical point evaluated at $f$. $\endgroup$ – user557493 Jun 13 '18 at 8:24
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    $\begingroup$ This is a different question. Please post it as such. I think that I've answered the original question. $\endgroup$ – José Carlos Santos Jun 13 '18 at 8:26
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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.

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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 $\pm\dfrac12$.

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  • $\begingroup$ Okay. So, the largest/smallest values of the local extrema (those on the boundary, stationary points etc) determine the global extrema? Hence if for the stationary point $(0,0)$, $f(0,0)\geq\frac{1}{2}$, then $(0,0)$ would be the global maximum? Is this what you mean? $\endgroup$ – user557493 Jun 13 '18 at 8:35
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    $\begingroup$ @Bell: right. Try with $1-x^2-y^2$ or $xy-x^2-y^2$. $\endgroup$ – Yves Daoust Jun 13 '18 at 8:37
  • $\begingroup$ I will indeed. That was very helpful. Thank you! $\endgroup$ – user557493 Jun 13 '18 at 8:38

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