use lagrange multipliers to find the maximum value of the function f(x,y,z)= x+2y+3z ?

First, set g1(x,y,z) = x-y+z-1 & g2(x,y,z) = x^2+y^2-1

Define the define the Lagrangian, Λ as

Λ(x,y,z,λ1,λ2) = f(x,y,z)+λ1*g1(x,y,z)+λ2*g2(x,y,z)

Thus,

Λ(x,y,z,λ1,λ2) = x+2y+3z+λ1*(x-y+z-1)+λ2*(x^2 + y^2-1)

Next, we require that the derivative dΛ = 0

This is equivalent to the problem of finding the solutions to:

(1) ∂Λ/∂x = 0

(2) ∂Λ/∂y = 0

(3) ∂Λ/∂z = 0

(4) ∂Λ/∂λ1 = 0

(5) ∂Λ/∂λ2 = 0

This leads to:

(1) 1+λ1+2λ2x = 0

(2) 2-λ1+2λ2y = 0

(3) 3 +λ1 = 0

(4) x-y+z-1 = 0

(5) x^2 + y^2-1 = 0

Clearly, (3) leads to λ1 = -3. Eliminate from the other equations to establish critical points.

Once you have the set of critical points (x1,y1,z1)....(xn,yn,zn) that are solutions to (1)-(5), evaluate f(x,y,z) at these points. The maximum value will be the global maximum (with respect to the constraints).