# How do you minimize and maximize f(x,y)=(x-2)^2-(y-3)^2/x constrained to 0<xy-y^2<5?

Feb 9, 2016

There are no maxima or minima of f, and no critical points on the set in question. There is only a saddle point and that's not in the constraint region.

#### Explanation:

The problem is that your constraint region is an open set , meaning it doesn't contain its boundary. Any constrained max and min is either at a critical point of f, on a constraint curve, or on the boundary of a constraint region.

There is a critical point of f at $\left(x , y\right) = \left(2 , 3\right)$, meaning the partials fx and fy are 0 there. But it's a saddle point and also not in the region because $x y - {y}^{2} = 2 \cdot 3 - {3}^{2} = - 3$, not between 0 and 5.

If instead you meant $0 \le x y - {y}^{2} \le 5$, then it's a whole different story. In that case the max and min of f along the curves $x y - {y}^{2} = 0$ and $x y - {y}^{2} = 5$ would be your answers.

Hope this helps!
// dansmath strikes again! \\