Question

How do you minimize and maximize `f(x,y)=x/y-xy` constrained to `0<x-y<1`?

Answer 1

`+-oo`

Explanation:

`lim_(x->oo)f(x,x-1/2)=lim_(x->oo)(x/(x-1/2)-x*(x-1/2))=-oo`
`lim_(y->0^+)f(y+1/2,y)=lim_(y->0^+)(y+1/2)/y-(y+1/2)y=+oo`

Both choices respect the constraint.

Answer 2

There is a local máxima at `x=0,y=-1`

Explanation:

Defining

`f(x,y)=x/y-x y`

and

`g_1(x,y,s)=x-y-s^2 = 0`
`g_2(x,y,s)=x-y-1+s^2=0`

the local maximization/minimization problem with inequality restrictions is transformed into an equivalent one, now with equality restrictions, so we can apply the Lagrange Multipliers technique for its resolution. The lagrangian is

`L(X,S,Lambda)=f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)`

Here `X = (x,y), S = (s_1,s_2), Lambda=(lambda_1,lambda_2)`

The stationary points are solutions of

`grad L =vec 0`

or

#{ (1/y - y + lambda_1 + lambda_2=0), ( x + x/y^2+ lambda_1 + lambda_2 = 0), (2 lambda_1 s_1 = 0), (2 lambda_2 s_2 = 0), ( x - y -s_1^2 = 0), ( x - y -1 + s_2^2= 0) :}#

Solving for `X,S,Lambda` we obtain

`(x = 0, y = -1, s_1 =pm1, s_2 = 0, lambda_1 = 0, lambda_2 = 0)`

Here we observe that `s_2=0` so the active restriction is

`g_2(x,y,0)=0`

Also we have

`(f @ g_2)(x)=1 + 1/(x-1) + x - x^2` and

`(d^2)/(dx^2)(f @ g_2)(x)=2/(x-1)^3-2` and
`(d^2)/(dx^2)(f @ g_2)(0)=-4`

This result shows that the point `x=0,y=-1` is a local máxima for the problem.