This implies that grad fis normal to the curve of intersection since it is perpendicular
to the tangent vector dx ̄ to the curve of intersection. At a stationary point, the
normal plane containing the vector grad falso contains the vectors ∇gand ∇hsince
dg =grad g·d ̄x= 0 and dh =grad h·dx = 0 at the stationary point. Hence, if these
three vectors are noncollinear, then there will exist scalars λ 1 and λ 2 such that
∇f+λ 1 ∇g+λ 2 ∇h= 0 (7 .82)
at a stationary point. The equation (7.82) is a vector equation and is equivalent to
the three scalar equations
∂f
∂x +λ^1
∂g
∂x +λ^2
∂h
∂x =0
∂f
∂y +λ^1
∂g
∂y +λ^2
∂h
∂y =0
∂f
∂z +λ^1
∂g
∂z +λ^2
∂h
∂z =0.
or
∂f
∂x 1
+λ 1 ∂g
∂x 1
+λ 2 ∂h
∂x 1
=0
∂f
∂x 2
+λ 1 ∂g
∂x 2
+λ 2 ∂h
∂x 2
=0
∂f
∂x 3 +λ^1
∂g
∂x 3 +λ^2
∂h
∂x 3 =0.
depending upon the notation you are using. These three equations together with
the constraint equations g= 0 and h= 0 gives us five equations in the five unknowns
x, y, z, λ 1 , λ 2 that must be satisfied at a stationary point.
By the Lagrangian rule one can form the function
F=F(x, y, z, λ 1 , λ 2 ) = f(x, y, z ) + λ 1 g(x, y, z) + λ 2 h(x, y, z)
where f(x, y, z ) is the objective function, g(x, y, z) and h(x, y, z ) are the constraint
functions and λ 1 , λ 2 are the Lagrange multipliers. Observe that F has a stationary
value where
∂F
∂x =
∂f
∂x +λ^1
∂g
∂x +λ^2
∂h
∂x = 0
∂F
∂y
=
∂f
∂y
+λ 1
∂g
∂y
+λ 2
∂h
∂y
= 0
∂F
∂z
=
∂f
∂z
+λ 1
∂g
∂z
+λ 2
∂h
∂z
= 0
∂F
∂λ 1 =g(x, y, z ) = 0
∂F
∂λ 2 =h(x, y, z) = 0
(7 .83)
These are the same five equations, with unknowns x, y, z, λ 1 , λ 2 ,for determining the
stationary points as previously noted.
