How do you minimize and maximize #f(x,y)=(x-y)/(x^3-y^3# constrained to #2<xy<4#?

1 Answer
Oct 13, 2016

See below.

Explanation:

How do you minimize and maximize #f(x,y)=(x-y)/(x^3-y^3)# constrained to #2 < xy <4 #?

Considering that #(x^3-y^3)/(x-y)=x^2+ xy+y^2# we have now the optimization problem

Find extrema of

#f(x,y) =1/(x^2+x y + y^2)#

subjected to

#g_1(x,y,s_1) = x y - 2 - s_1^2 =0#
#g_2(x,y,s_2) = x y - 4 + s_2^2=0#

#s_1,s_2# are known as slack variables. They are used to transform the inequality into equality relations.

The lagrangian is

#L(x,y,s_1,s_2,lambda_1,lambda_2)=f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)# and it's stationary points are the points obeying

#grad L = vec 0# or

#{ (lambda_1 y + lambda_2 y - (2 x + y)/(x^2 + x y + y^2)^2 = 0), (lambda_1 x + lambda_2 x - (x + 2 y)/(x^2 + x y + y^2)^2 = 0), (-2 lambda_1 s_1 = 0), (2 lambda_2 s_2 = 0), (s_1^2 + x y - 2= 0), (s_2^2 + x y - 4= 0):}#

Solving for #x,y,s_1,s_2,lambda_1,lambda_2# we obtain
#p_1,p_2,p_3,p_4#

#( (x = -2., y = -2., s_1 = -1.41421, s_2 = 0, lambda_1 = 0, lambda_2 -> 0.0208333), (x = 2., y= 2., s_1 = -1.41421, s_2 = 0, lambda_1= 0, lambda_2 = 0.0208333), (x = -1.41421, y= -1.41421, s_1 = 0, s_2 = -1.41421, lambda_1 =0.0833333, lambda_2 = 0), (x = 1.41421, y = 1.41421, s_1 = 0, s_2= -1.41421, lambda_1= 0.0833333, lambda_2 = 0))#

The stationary points qualification is done according to the sign of

#(d^2(f@g_1)(s_1=0))/(dx^2)# and
#(d^2(f@g_2)(s_2=0))/(dx^2)#

Here

#(f@g_1)(x) = 1/(2 + 4/x^2 + x^2)#
#(f@g_2)(x) = 1/(4 + 16/x^2 + x^2)#

and

#(d^2(f@g_1)(s_1=0))/(dx^2) < 0# and
#(d^2(f@g_2)(s_2=0))/(dx^2) < 0# for all points so

#p_1,p_2,p_3,p_4# are all local maxima.

Attached a plot showing the feasible region and the points location.

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