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In the case of an optimization function with three variables and a single constraint function, it is possible to use the method of Lagrange multipliers to solve an optimization problem as well. An example of an optimization function with three variables could be the Cobb-Douglas function in the previous example: f ( x , y , z ) = x 0.2 y 0.4 z 0.4 , where x represents the cost of labor, y represents capital input, and z represents the cost of advertising. The method is the same as for the method with a function of two variables; the equations to be solved are

f ( x , y , z ) = λ g ( x , y , z ) g ( x , y , z ) = 0.

Lagrange multipliers with a three-variable optimization function

Maximize the function f ( x , y , z ) = x 2 + y 2 + z 2 subject to the constraint x + y + z = 1 .

  1. The optimization function is f ( x , y , z ) = x 2 + y 2 + z 2 . To determine the constraint function, we subtract 1 from each side of the constraint: x + y + z 1 = 0 which gives the constraint function as g ( x , y , z ) = x + y + z 1 .
  2. Next, we calculate f ( x , y , z ) and g ( x , y , z ) :
    f ( x , y , z ) = 2 x , 2 y , 2 z g ( x , y , z ) = 1 , 1 , 1 .

    This leads to the equations
    2 x 0 , 2 y 0 , 2 z 0 = λ 1 , 1 , 1 x 0 + y 0 + z 0 1 = 0

    which can be rewritten in the following form:
    2 x 0 = λ 2 y 0 = λ 2 z 0 = λ x 0 + y 0 + z 0 1 = 0.
  3. Since each of the first three equations has λ on the right-hand side, we know that 2 x 0 = 2 y 0 = 2 z 0 and all three variables are equal to each other. Substituting y 0 = x 0 and z 0 = x 0 into the last equation yields 3 x 0 1 = 0 , so x 0 = 1 3 and y 0 = 1 3 and z 0 = 1 3 which corresponds to a critical point on the constraint curve.
  4. Then, we evaluate f at the point ( 1 3 , 1 3 , 1 3 ) :
    f ( 1 3 , 1 3 , 1 3 ) = ( 1 3 ) 2 + ( 1 3 ) 2 + ( 1 3 ) 2 = 3 9 = 1 3 .

    Therefore, an extremum of the function is 1 3 . To verify it is a minimum, choose other points that satisfy the constraint and calculate f at that point. For example,
    f ( 1 , 0 , 0 ) = 1 2 + 0 2 + 0 2 = 1 f ( 0 , −2 , 3 ) = 0 2 + + ( −2 ) 2 + 3 2 = 13.

    Both of these values are greater than 1 3 , leading us to believe the extremum is a minimum.
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Use the method of Lagrange multipliers to find the minimum value of the function

f ( x , y , z ) = x + y + z

subject to the constraint x 2 + y 2 + z 2 = 1 .

f ( 3 3 , 3 3 , 3 3 ) = 3 3 + 3 3 + 3 3 = 3 f ( 3 3 , 3 3 , 3 3 ) = 3 3 3 3 3 3 = 3 .

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Problems with two constraints

The method of Lagrange multipliers can be applied to problems with more than one constraint. In this case the optimization function, w is a function of three variables:

w = f ( x , y , z )

and it is subject to two constraints:

g ( x , y , z ) = 0 and h ( x , y , z ) = 0 .

There are two Lagrange multipliers, λ 1 and λ 2 , and the system of equations becomes

f ( x 0 , y 0 , z 0 ) = λ 1 g ( x 0 , y 0 , z 0 ) + λ 2 h ( x 0 , y 0 , z 0 ) g ( x 0 , y 0 , z 0 ) = 0 h ( x 0 , y 0 , z 0 ) = 0.

Lagrange multipliers with two constraints

Find the maximum and minimum values of the function

f ( x , y , z ) = x 2 + y 2 + z 2

subject to the constraints z 2 = x 2 + y 2 and x + y z + 1 = 0 .

Let’s follow the problem-solving strategy:

  1. The optimization function is f ( x , y , z ) = x 2 + y 2 + z 2 . To determine the constraint functions, we first subtract z 2 from both sides of the first constraint, which gives x 2 + y 2 z 2 = 0 , so g ( x , y , z ) = x 2 + y 2 z 2 . The second constraint function is h ( x , y , z ) = x + y z + 1 .
  2. We then calculate the gradients of f , g , and h :
    f ( x , y , z ) = 2 x i + 2 y j + 2 z k g ( x , y , z ) = 2 x i + 2 y j 2 z k h ( x , y , z ) = i + j k .

    The equation f ( x 0 , y 0 , z 0 ) = λ 1 g ( x 0 , y 0 , z 0 ) + λ 2 h ( x 0 , y 0 , z 0 ) becomes
    2 x 0 i + 2 y 0 j + 2 z 0 k = λ 1 ( 2 x 0 i + 2 y 0 j 2 z 0 k ) + λ 2 ( i + j k ) ,

    which can be rewritten as
    2 x 0 i + 2 y 0 j + 2 z 0 k = ( 2 λ 1 x 0 + λ 2 ) i + ( 2 λ 1 y 0 + λ 2 ) j ( 2 λ 1 z 0 + λ 2 ) k .

    Next, we set the coefficients of i and j equal to each other:
    2 x 0 = 2 λ 1 x 0 + λ 2 2 y 0 = 2 λ 1 y 0 + λ 2 2 z 0 = −2 λ 1 z 0 λ 2 .

    The two equations that arise from the constraints are z 0 2 = x 0 2 + y 0 2 and x 0 + y 0 z 0 + 1 = 0 . Combining these equations with the previous three equations gives
    2 x 0 = 2 λ 1 x 0 + λ 2 2 y 0 = 2 λ 1 y 0 + λ 2 2 z 0 = −2 λ 1 z 0 λ 2 z 0 2 = x 0 2 + y 0 2 x 0 + y 0 z 0 + 1 = 0.
  3. The first three equations contain the variable λ 2 . Solving the third equation for λ 2 and replacing into the first and second equations reduces the number of equations to four:
    2 x 0 = 2 λ 1 x 0 2 λ 1 z 0 2 z 0 2 y 0 = 2 λ 1 y 0 2 λ 1 z 0 2 z 0 z 0 2 = x 0 2 + y 0 2 x 0 + y 0 z 0 + 1 = 0.

    Next, we solve the first and second equation for λ 1 . The first equation gives λ 1 = x 0 + z 0 x 0 z 0 , the second equation gives λ 1 = y 0 + z 0 y 0 z 0 . We set the right-hand side of each equation equal to each other and cross-multiply:
    x 0 + z 0 x 0 z 0 = y 0 + z 0 y 0 z 0 ( x 0 + z 0 ) ( y 0 z 0 ) = ( x 0 z 0 ) ( y 0 + z 0 ) x 0 y 0 x 0 z 0 + y 0 z 0 z 0 2 = x 0 y 0 + x 0 z 0 y 0 z 0 z 0 2 2 y 0 z 0 2 x 0 z 0 = 0 2 z 0 ( y 0 x 0 ) = 0. .

    Therefore, either z 0 = 0 or y 0 = x 0 . If z 0 = 0 , then the first constraint becomes 0 = x 0 2 + y 0 2 . The only real solution to this equation is x 0 = 0 and y 0 = 0 , which gives the ordered triple ( 0 , 0 , 0 ) . This point does not satisfy the second constraint, so it is not a solution.
    Next, we consider y 0 = x 0 , which reduces the number of equations to three:
    y 0 = x 0 z 0 2 = x 0 2 + y 0 2 x 0 + y 0 z 0 + 1 = 0.

    We substitute the first equation into the second and third equations:
    z 0 2 = x 0 2 + x 0 2 x 0 + x 0 z 0 + 1 = 0.

    Then, we solve the second equation for z 0 , which gives z 0 = 2 x 0 + 1 . We then substitute this into the first equation,
    z 0 2 = 2 x 0 2 ( 2 x 0 + 1 ) 2 = 2 x 0 2 4 x 0 2 + 4 x 0 + 1 = 2 x 0 2 2 x 0 2 + 4 x 0 + 1 = 0 ,

    and use the quadratic formula to solve for x 0 :
    x 0 = −4 ± 4 2 4 ( 2 ) ( 1 ) 2 ( 2 ) = −4 ± 8 4 = −4 ± 2 2 4 = −1 ± 2 2 .

    Recall y 0 = x 0 , so this solves for y 0 as well. Then, z 0 = 2 x 0 + 1 , so
    z 0 = 2 x 0 + 1 = 2 ( −1 ± 2 2 ) + 1 = −2 + 1 ± 2 = −1 ± 2 .

    Therefore, there are two ordered triplet solutions:
    ( −1 + 2 2 , −1 + 2 2 , −1 + 2 ) and ( −1 2 2 , −1 2 2 , −1 2 ) .
  4. We substitute ( −1 + 2 2 , −1 + 2 2 , −1 + 2 ) into f ( x , y , z ) = x 2 + y 2 + z 2 , which gives
    f ( −1 + 2 2 , −1 + 2 2 , −1 + 2 ) = ( −1 + 2 2 ) 2 + ( −1 + 2 2 ) 2 + ( −1 + 2 ) 2 = ( 1 2 + 1 2 ) + ( 1 2 + 1 2 ) + ( 1 2 2 + 2 ) = 6 4 2 .

    Then, we substitute ( −1 2 2 , −1 2 2 , −1 2 ) into f ( x , y , z ) = x 2 + y 2 + z 2 , which gives
    f ( −1 2 2 , −1 2 2 , −1 2 ) = ( −1 2 2 ) 2 + ( −1 2 2 ) 2 + ( −1 2 ) 2 = ( 1 + 2 + 1 2 ) + ( 1 + 2 + 1 2 ) + ( 1 + 2 2 + 2 ) = 6 + 4 2 .

    6 + 4 2 is the maximum value and 6 4 2 is the minimum value of f ( x , y , z ) , subject to the given constraints.
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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