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

Question

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

Isabella Ackerman · Accepted 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.

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! // dansmath strikes again! \

Emilia Aguilar · Answer

undefined To minimize and maximize $ f(x,y) = \frac{{(x-2)^2 - (y-3)^2}}{{x}} $ constrained to $ 0 < xy - y^2 < 5 $, you first need to find the critical points of $ f(x,y) $ within the given constraint. Then, evaluate $ f(x,y) $ at these critical points as well as at the boundary points of the constraint region to determine the minimum and maximum values of $ f(x,y) $.

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

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