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_1s1 and s_2s2 such that the problem will read.
Maximize/minimize f(x,y) = x e^x - yf(x,y)=xex−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