Lagrange Multipliers

(1) Proposition (Lagrange's Theorem) Assume that and have continuous first partial derivatives and that has an extremum at on the smooth constraint curve If there is a number such that

    Suppose and satisfy the hypotheses of Lagrange's theorem, and that has an extremum subject to the constraint Then to find the extreme value, proceed as follows:

(i)  Simultaneously solve the following three equations for and


(ii)  Evaluate at all points found in step (i). The extremum we seek must be among these values.

(2) Example (Lagrange's Theorem)  Use the method of Lagrange multipliers to find the required constrained extrema.

(a) Find the extreme values of the function on the circle

    Solution. Using Lagrange multipliers, we solve the equations which can be written as

or

From the first equation we have or If then by the third equation If then and we obtain Therefore, has possible extreme values at the points and Evaluating at these four points, we find that and

Therefore the maximum value of on the circle is and the minimum value is

lagrange multipliers _gr_44.gif]

(b) Maximize subject to

    Solution. Let then we have and We need to solve the system and We find that Therefore, is the constrained maximum.

(c) Minimize subject to

    Solution. Let then we have    and We need to solve the system and We find that and then and Therefore, is the constrained minimum.

(d) A rectangular box with no top is to be constructed from 96 of material. What should be the dimensions of the box if it is to enclose maximum volume?

    Solution. Let and be the length, width, and height of the rectangular box, respectively. We want to maximize the volume: subject to which is obtained from the surface area of the rectangular box (with no lid). We have Solve the system

We obtain and then find Therefore the maximum volume is

(e) A cylindrical can is to hold of orange juice.  The cost per square inch of constructing the metal top and bottom is twice the cost per square inch of constructing the cardboard side. What are the dimensions of the least expensive can?

    Solution. Let and be the radius and height of the cylinder, respectively. We want to minimize the cost subject to the constraint where We have and Solving the system

we obtain and then find the radius in. and the height in.

Suppose is an extreme value of subject to the constraint Then the Lagrange multiplier is the rate of change of with respect to ; that is Note that at the extreme value we have

The coordinates of the optimal ordered pair depend on (because different constraint levels will generally lead to different optimal combinations of and ). Thus, where and are functions of By the chain rule for partial derivatives:

(3) Example (Lagrange Multipliers With Two Parameters) Use the method of Lagrange multipliers to find the required extrema for the two given constraints.

(a)  Find the maximum value of the function

on the curve of intersection of the plane and the cylinder

    Solution. We maximize the function

subject to the constraints

The Lagrange condition is so we solve the equations

lagrange multipliers _gr_135.gif]

Putting we get so Similarly, we have Substitution yields and so Then and so we have The corresponding values of are

Therefore the maximum of on the given curve is

(b) Find the maximum of subject to and

    Solution. We need to solve the system where and Therefore, we need to solve the system



which is

The solutions are

The maximum is

(c)  Find the minimum of subject to and

    Solution. We want to minimize subject to the side conditions

We form

The conditions are












The first and third conditions give

so the second condition becomes We then have

lagrange multipliers _gr_182.gif]

The solution to this system is Therefore, the minimum is

(d) Use Lagrange multipliers to find the point on the line of intersection of the planes and that is closest to the origin.

    Solution. We want to minimize subject to the side conditions

We form

The conditions are












The second and third conditions give

so the first condition becomes We then have

lagrange multipliers _gr_197.gif]

The last two equations may be written as and Substitution of these values into the first equation gives Consequently, and The desired point is therefore,

Cite this as:
Lagrange Multipliers
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
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