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

1 Answer
Feb 9, 2016

Answer:

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 #(x,y) = (2,3)#, 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*3 - 3^2 = -3#, not between 0 and 5.

If instead you meant #0<=x y - y^2<=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!
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