A simple contrained optimization problem

Question

Suppose we have to find out the solution of the following optimization problem $$ {\min \atop x\,y =1}{x^2+y^2 \atop } $$ Clearly, in this case, we can apply the classical graphical method of finding where the level curves of the function $f(x,y)=x^2+y^2$ are tangent to the constraint $x\,y=1$. On the other hand if we want to solve $$ {\max \atop x\,y =1}{x^2+y^2 \atop } $$ it looks like the solutions are $x=+\infty$ and $y=0$ or $x=0$ and $y=+\infty$, in the sense that moving toward any of these two directions will provide a better solution than that given by the classical graphical method (i.e. $(x,y)=(1,1)$). So when the graphical method is not applicable to this kind of problems? In standard textbooks look like it is given for granted (i.e. it is typically said that the curve of the constraint must be tangent and not cross the level curve), so probably I am missing something or my reasoning is flawed and I do not see it.

Answer 1

For the minimization problem you can take $y=\dfrac{1}{x}$ and subs on $x^2+y^2$ to get $x^2+\dfrac{1}{x^2}$ on which you can use elementary calculus (its derivative and equate to zero) to verify that the minimum are at $(1,1)$ and $(-1,-1)$.

There is not maximum because $x^2+\dfrac{1}{x^2}$ can be large as $x\to\infty$.

Answer 2

Normally an optimization is meant to find a maximum or a minimum (on a "well behaved" function) restricted on some subset. In your second example the "solution" you provide is not a maximum or a minimum as it is not a stationary point (where the second derivative is zero). This is a necessary (but not sufficient) condition.

In any case, there is no maximum for your problem. You can always find a larger value of $x^2+y^2$ constrained for $xy=1$.