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

2 Answers
Nov 7, 2016

+-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.

Nov 7, 2016

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.