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

Given #f(x,y)=x^2-y/x# subjected to #Omega(x,y)#

#Omega(x,y) ={ (x,y) | 0 < x+3y < 2}#

We are looking for local minima/maxima.

This problem can be handled using lagrange multipliers. This can be done following the steps:

1) Describe #Omega(x,y)# through equality constraints
This can be done introducing the so called slack variables #s_1,s_2# and making

#g_1(x,y,s_1) = x + 3y - s_1^2=0#
#g_2(x,y,s_2) = x + 3y - 2+ s_2^2=0#

then #Omega(x,y) equiv g_1(x,y,s_1) nn g_2(x,y,s_2)#

2) Form the Lagrangian

#L(x,y,s_1,s_2,lambda_1,lambda_2)=f(x,y)+lambda_1g_1(x,y,s_1)+lambda_2 g_2(x,y,s_2)#

3) Determine the stationary points of #L(x,y,s_1,s_2,lambda_1,lambda_2)#

This is done computing the solution to

#grad L(x,y,s_1,s_2,lambda_1,lambda_2) = vec 0#

where #grad# represents the partial derivatives operator

#grad = (partial/(partial x),partial/(partial y),partial/(partial s_1),partial/(partial s_2),partial/(partial lambda_1),partial/(partial lambda_2))#

so the equations which define the stationary points are

#{ (lambda_1 + lambda_2 + 2 x + y/x^2= 0), (3 lambda_1 + 3 lambda_2 -1/x= 0), (2 lambda_1 s_1 =0), (2 lambda_2 s_2 = 0), ( x + 3 y-s_1^2 = 0), (x + 3 y-2 + s_2^2 = 0):}#

Solving for #x,y,s_1,s_2,lambda_1,lambda_2# we obtain the only real solution

#((f(x,y),x,y,s_1,s_2,lambda_1,lambda_2),(1.775,-0.693, 0.897., -1.414, 0,0, -0.480) )#

4) Qualifying the stationary points

This can be done computing #(d^2(f @ g_i))/(dx^2)# for #i=1,2#

and depending on the sign, if positive it is a local minimum and if negative a local maximum.

In our case,

#(f@g_1)(x) = 1/3 + x^2#
#(f@g_2)(x) =1/3 - 2/(3 x) + x^2#

#(d^2)/(dx^2)(f@g_1)(x) = 2#
#(d^2)/(dx^2)(f@g_2)(x) =2 - 4/(3 x^3)#

The evaluations must be done according to the values found for #s_1# and #s_2# so, this point shall be qualified with #(d^2)/(dx^2)(f@g_2)(x), (s_2= 0)#

Attached a plot showing the region with the objective function level curves, and the stationary point.

The qualification is left to the reader. I think it is a local minimum.

Note. Of course if the evaluation gives zero we will continue the qualification process but this is another chapter.

Maximization Note that the constraint #0 < x+3y <2# allows #xrarr +oo# (provided #3y rarr -oo#) In which case #f(x,y) rarr +oo#

Minimization Note that the constraint # 0 < x+3y < 2# allows #x rarr 0+# (provided #y > 0#) In which case #f(x,y) rarr -oo#