Lagrange Multipliers With One Parameter

    In many applied problems, the main focus is on optimizing a function subject to constraint; for example, finding extreme values of a function of several variables where the domain is restricted to a level curve (or surface) of another function of several variables. Lagrange multipliers is a general method which can be used to solve such optimization problems.

Lagrange Multipliers With One Parameter

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



    Proof. Denote the constraint curve by and note that is smooth. We represent this curve by the vector function for all in an open interval including corresponding to where and exist and are continuous. Let for all in and apply the chain rule to obtain



Because has an extremum at we know that has an extremum at Therefore, we have and



If then and the condition   is satisfied trivially. If then is orthogonal to Because is tangent to the constraint curve , it follows that is normal to But is also normal to (because is a level curve of ), and we conclude that and must be parallel at Thus, there is a scalar 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.
    

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



(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:

• print this page
• pages linking here
• related pages

Cite this as:
Lagrange Multipliers With One Parameter
Published by Library of Math -- Online math organized by subject into topics.
Written by Smith, David A.
http://www.libraryofmath.com/lagrange-multipliers-with-one-parameter.html