Calc 3 Lecture Notes Section 12.8 Page 8 of 9

Section 12.8: Constrained Optimization and Lagrange Multipliers

Big idea: The fact that a function has an absolute extremum on a closed, bounded region can be combined with the geometric interpretation of the gradient to quickly find an extremum of a function when it is constrained by another function.

Big skill: You should be able to use Lagrange multipliers to optimize a function subject to a constraint.

Warm-up problem:

Find the point on the line y = 3 – 2x that is closest to the origin.

Using the analytic geometry technique that the shortest distance from a point to a line is along the perpendicular to the line:


Using the Calc 1 technique of substituting the line’s equation into the square of the distance formula and then minimizing the resulting function of one variable:


The Calc 3 technique for solving this problem is to minimize the function for the square of the distance subject to the constraint that the point must lie on the line described by . At the solution point, the level curve of the function f will be tangent to the level curve of the function g. Thus, their gradients will be parallel at that point, and so:

.


Theorem 8.1:

If and are functions with continuous first partial derivatives and on the surface and either the minimum or maximum value of subject to the constraint occurs at , then

for some constant l (called the Lagrange multiplier).

This technique comes from imagining that the function f has an extremum at (x0, y0, z0) lying on the level surface S defined by g = 0. We can construct any arbitrary curve defined parametrically that lies on S and passes through (x0, y0, z0) with a vector-valued function r(t) = <x(t), y(t), z(t)>, and then substituting this parameterization into f to form a new function h(t). Working out the implication of h¢(t0) = 0 at the extremum, it can be seen that Ñf is always orthogonal to any and all tangent vectors r¢. Since grad g is also orthogonal to S, we get that Ñf and Ñg must be parallel.

Notice that Theorem 8.1 implies a system of four equations in four variables to be solved:

When Theorem 8.1 is applied to functions of two variables, it implies solving a system of three equations in three unknowns:

Practice:

  1. Find the extreme temperatures on a metal plate with temperature function
    T(x, y) = x2 + 2x + y2 that is in the shape of x2 + 4y2 £ 24.


Practice: Solve the following three-variable constraint problem:

  1. The profit a company makes on producing x, y, and z thousand units of products is given by: P(x, y, z) = 4x + 8y + 6z. If manufacturing constraints force . Find the production parameters for maximum profit for the company.


Optimization with two constraints:

To optimize the function subject to the constraints and , solve

Notice that two constraints on a function of three variables implies a system of five equations in five variables to be solved:

Alternative geometric calculation: find the curve of intersection for and , then find extrema of f along that curve.


Practice:

  1. The plane x + y + z = 12 and the paraboloid z = x2 + y2intersect in an ellipse. Find the point on that ellipse closest to the origin.