Read the article in Lagrange multiplier.And also find the value of x,y,z with help given book. According to the book "Sparse and Redundant Representations: From Theory to Applications in Signal and Image Processing" by Michael Elad, in the section of "Regularization" (p.4), or read the article in Lagrange multiplier 
try to solve the optimization problem with
        f(x,y)=3−(2x^2+y^2)

          g(x,y)=x+y=3,

and then obtain 
∂L/∂x , ∂L/∂y , ∂L/∂λ
and the solution of x,y,z.
 A: Given equation:
    f ( x , y )   =    3 +  ( 2x^2 + y^2 )
    g ( x , y )  =   x + y = 3 ,
Lagrange multiplier defined by
L( x , y , λ  ) = f( x , y ) - λ g(x,y)

where the λ term may be either added or subtracted. If f(x0, y0) is a maximum of f(x, y) for the original constrained problem, then there exists λ0 such that (x0, y0, λ0) is a stationary point for the Lagrange function (stationary points are those points where the partial derivatives of L are zero).
L(x) = 3−(2x^2+y^2) + λ(x+y-3);
With λ being the Lagrange multiplier for the constrain set .Taking a derivative of L(x) with respect to x ,we obtain the requirement
∂L(x)/∂x = 0
 -4x+ λ = 0                 .............( 1 )
∂L(x)/∂y = 0
-2y+λ = 0                    .............( 2 )
∂L(x)/∂λ = 0
x+y = 0                       .............( 3 )
Substrate the equation (1) & (2), we get 
-2x+y = 0                  ...............(4)
Again substrate the equation (3) & (4),we get
3x =3
x=1;
Putting the value of x in the equation (3),
We get 
1+ y = 3
y=2;
Again Putting the value of x and y in the equation (1) ,we get 
-4(1) + λ =0
λ = 4;
Answer :x=1; y=2 and λ=4
