# Find max/min of the following function

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?

• 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. – copper.hat Jan 27 at 17:17
• Why the fact that the set C is convex and closed implies that there is a unique min point ? – LearningProb Jan 27 at 17:28
• 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. – bjorn93 Jan 27 at 17:31
• 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.) – copper.hat Jan 27 at 17:32
• Now it clear thank you. – LearningProb Jan 27 at 17:57

(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.

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

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?

• Hi Emilio, yes I can do this but then how do I prove that the critical points are actually minima? – LearningProb Jan 27 at 18:35
• I got that $(2,1/2)$ is a critical but how do I prove that it is a global min? – LearningProb Jan 27 at 21:49
• Use the Hessian test (very simple in this case): en.wikipedia.org/wiki/Hessian_matrix – Emilio Novati Jan 27 at 22:47
• 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. – LearningProb Jan 28 at 1:09
• Hi, I can not find it – LearningProb Jan 28 at 12:21

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 $$\sqrt{\frac{17}{4}}$$