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Find the minimum e max distance (in $R^2$) ) between the point $Q = ( 3/ 2 , − 3/ 2 )$ and the set $$B = \{(x, y) ∈ R^2 : yx = 1, x ≥ 0, y ≥ 0\}$$

In other words I have to find max /min points of the function

$${(x-3/2)^2 + (y+3/2)^2}$$

The set $B$ is clearly neither bounded nor convex. If I use Lagrange I can only find local min/max, but how do I show they are global? In a previous question :Does the following function admit a maximum? you suggested me to maximize/minimize the $x$ component and the $y$ component independently, but it is not clear to me if I can do it in this exercise as well. It seens to me that choosing $x=3/2$, which clearly minimizes the first component of the sum, would restrict the choice of $y$ as $yx = 1 $.

Is there any way to show that the point found using lagrange is a global min? Possibly without using the bordered hessian method?

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  • $\begingroup$ Note that $C=\{ (x,y) | yx \ge 1, x,y \ge 0 \}$ is convex (and closed) and $Q \notin C$, so there is a unique point of minimal norm. Since the minimiser must be on the boundary and $B$ is the boundary, it is the minimiser on $B$ and hence is unique. $\endgroup$ – copper.hat Jan 27 at 17:17
  • $\begingroup$ Why the fact that the set C is convex and closed implies that there is a unique min point ? $\endgroup$ – LearningProb Jan 27 at 17:28
  • $\begingroup$ If the point is $(3/2,-3/2)$, then the distance is $\dots+(y+3/2)^2$. The problem can be solved directly by writing it as a function of $x$ and taking 1st derivative. I'm not sure if that's what you want. $\endgroup$ – bjorn93 Jan 27 at 17:31
  • $\begingroup$ Well, since $A=\overline{B}(Q, 100)$ (say) is closed and bounded and must contain any closest point we can consider the minimum of a convex function (distance) over a convex compact set and a minimiser must exist. (It is a more general result that a closed convex set has a unique nearest point in the Euclidean norm.) $\endgroup$ – copper.hat Jan 27 at 17:32
  • $\begingroup$ Now it clear thank you. $\endgroup$ – LearningProb Jan 27 at 17:57
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(Note that you have a typo in your function, if the point is $(3/2,-3/2)$ then the second minus in your function should be a plus)

Using Lagrange multipliers is overkill here. In any case, Lagrange multipliers will give you critical points, which then you can check to see if they are the actual max/min.

Here, $x,y>0$ (otherwise $xy\ne1$), and you have that $y=1/x$. So your function becomes $$ \left(x-\tfrac32\right)^2+\left(\tfrac1x+\tfrac32\right)^2,\ \ \ \ x>0. $$ Now you can do this using one-variable calculus.

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  • $\begingroup$ Is the equation $f'(x,y)=0$ more easy to solve than the Lagrange system? $\endgroup$ – Emilio Novati Jan 27 at 18:10
  • $\begingroup$ As far as I can tell, you end up getting the exact same quartic, but it has two easy roots. $\endgroup$ – Martin Argerami Jan 27 at 18:11
  • $\begingroup$ The equation $f'=0$ is of degree 3, but the system is of degree 2 (I think). Anyway, yes, there is the simple solution $x=2$. $\endgroup$ – Emilio Novati Jan 27 at 18:13
  • $\begingroup$ Yes, you are right. What happened is that I solved super quickly just to check and I used a non-efficient idea that increased the degree. $\endgroup$ – Martin Argerami Jan 27 at 18:32
  • $\begingroup$ Using both lagrange and $f'=0$ I still get $x^4$ but collecting terms it can then be simplified a little bit. $\endgroup$ – LearningProb Jan 27 at 18:40
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Hint:

The square of the distance function that you want minimize is $f(x,y)=(x-\frac{3}{2})^2+(y+\frac{3}{2})^2$ ( it seems that you have a wrong sign) with the condition $g(xy)=xy=1$ so, using Lagrange multipliers, you have to solve; $$ \begin{cases} \nabla f=\lambda \nabla g\\ xy=1 \end{cases} $$

can you do this?

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  • $\begingroup$ Hi Emilio, yes I can do this but then how do I prove that the critical points are actually minima? $\endgroup$ – LearningProb Jan 27 at 18:35
  • $\begingroup$ I got that $(2,1/2)$ is a critical but how do I prove that it is a global min? $\endgroup$ – LearningProb Jan 27 at 21:49
  • $\begingroup$ Use the Hessian test (very simple in this case): en.wikipedia.org/wiki/Hessian_matrix $\endgroup$ – Emilio Novati Jan 27 at 22:47
  • $\begingroup$ Do you mean the bordered hessian method? Actually I do not know that method, that it is why I was looking for an alternative to it. $\endgroup$ – LearningProb Jan 28 at 1:09
  • $\begingroup$ Hi, I can not find it $\endgroup$ – LearningProb Jan 28 at 12:21
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Lets say y->0 and x->$\infty$ then value of above expression -> $\infty$ (so yeah no maximum).

$(x-\frac{3}{2})^2 +(y+\frac{3}{2})^2$(You have made a mistake here or given the point wrong) = $x^2 +y^2 +\frac{9}{2} + 3(y-x) = z^2 +3z +\frac{13}{2}$.

(as $x=\frac{1}{y}$, $(y-\frac{1}{y})$=z, z can be any real number , the quadratic equation have its minimum at $\frac{-3}{2}$ which is the required answer)

So minimum value is $\sqrt{\frac{17}{4}}$

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