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Lagrange Multipliers and Constrained Optimization

In a Nut Shell: Sometimes you wish to maximize or minimize a function subject to one

or more limitations (called constraints). For example, one may wish to find the maximum

volume of an object subject to a restriction on its surface area or you may wish to maximize

profit subject to a restriction on available investment opportunities. The use of

Lagrange multipliers offers a convenient approach. Three cases will be discussed below.

Case 1 Suppose you wish to maximize/minimize a function with two independent

variables, f(x, y), subject to one constraint, g(x, y) .

The functions involved are:

f(x, y) = 0 and g(x, y) = 0

Let f(x,y) be the function to be optimized subject to the constraint relation, g(x,y).

Both f(x, y) and g(x, y) must be continuously differentiable functions.

Then introduce an arbitrary constant, λ , called the Lagrange multiplier such that:

Grad f = λ Grad g ------------------------------------ ( 1 )

where Grad g means gradient of g

So ∂f/∂x i + ∂f/∂y j = λ [∂g/∂x i + ∂g/∂y j ]

In scalar form: ∂f/∂x = λ [∂g/∂x ], ∂f/∂y = λ [ ∂g/∂y ], and g(x, y) = 0

Which gives 3 equations in the 3 unknowns, x, y, and λ

Once solved, the values of x and y can be input to f(x, y) to obtain the optimum.

Case 2 A similar approach applies to functions with three independent variables

subject to one constraint. i.e. The result is 4 equations in the 4 unknowns, x, y, z, and λ

The functions involved are:

f(x, y, z) = 0 subject to the constraint g(x, y, z) = 0

∂f/∂x = λ [∂g/∂x ], ∂f/∂y = λ [ ∂g/∂y ],

∂f/∂z = λ [ ∂g/∂z ], and g(x, y, z) = 0

Click here to continue with Case 3.

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