Find max/min of the following function

Question

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?

Answer 1

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

Answer 2

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?

Answer 3

(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.