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.