We will searching for stationary points, qualifying then as local maxima/minima.
First we will transform the maxima/minima with inequality restrictions into an equivalent maxima/minima problem but now with equality restrictions.
To do that we will introduce the so called slack variables #s_1# and #s_2# such that the problem will read.
Maximize/minimize #f(x,y) = x e^x - y#
constrained to
#{
(g_1(x,y,s_1)=x - y - s_1^2=0),
(g_2(x,y,s_2)=x - y + s_2^2 - 1=0)
:}#
The lagrangian is given by
#L(x,y,s_1,s_2,lambda_1,lambda_2) = f(x,y)+lambda_1 g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)#
The condition for stationary points is
#grad L(x,y,s_1,s_2,lambda_1,lambda_2)=vec 0#
so we get the conditions
#{
(e^x + lambda_1 + lambda_2 + e^x x = 0),
( -1 - lambda_1 - lambda_2 = 0),
(-s_1^2 + x - y = 0),
( -2 lambda_1 s_1 = 0),
( -1 + s_2^2 + x - y = 0),
( 2 lambda_2 s_2 = 0)
:}#
Solving for #{x,y,s_1,s_2,lambda_1,lambda_2}# we have
#{(x =0, y = -1, lambda_1 = 0., s_1 = 1,
lambda_2 = -1, s_2 = 0.), (x = 0, y = 0, lambda_1 =-1., s_1 = 0, lambda_2 = 0,
s_2 = 1.)
:}#
so we have two points #p_1={0,-1}# and #p_2 = {0,0}#
Point #p_1# activates restriction #g_2(x,y,0) = 0,{lambda_2 ne 0, s_2 = 0}# and point #p_2# activates restriction #g_1(x,y,0)=0,{lambda_1 ne 0, s_1 = 0}#
#p_1# is qualified with #f_{g_2}(x) =1 + (e^x-1) x#
and
#p_2# is qualified with #f_{g_1}(x) = (e^x-1) x#
Computing
#d/(dx)(f_{g_2}(0)) = 0#
and
#d^2/(dx^2)(f_{g_2}(0)) = 2#
we conclude that #p_1# local minimum point.
Analogously for #p_2#
#d/(dx)(f_{g_1}(0)) = 0#
and
#d^2/(dx^2)(f_{g_1}(0)) = 2#
so #p_1,p_2# are local minima points
