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If the boundary of the set D is a more complicated curve defined by a function g ( x , y ) = c for some constant c , and the first-order partial derivatives of g exist, then the method of Lagrange multipliers can prove useful for determining the extrema of f on the boundary. The method of Lagrange multipliers is introduced in Lagrange Multipliers .

Finding absolute extrema

Use the problem-solving strategy for finding absolute extrema of a function to determine the absolute extrema of each of the following functions:

  1. f ( x , y ) = x 2 2 x y + 4 y 2 4 x 2 y + 24 on the domain defined by 0 x 4 and 0 y 2
  2. g ( x , y ) = x 2 + y 2 + 4 x 6 y on the domain defined by x 2 + y 2 16
  1. Using the problem-solving strategy, step 1 involves finding the critical points of f on its domain. Therefore, we first calculate f x ( x , y ) and f y ( x , y ) , then set them each equal to zero:
    f x ( x , y ) = 2 x 2 y 4 f y ( x , y ) = −2 x + 8 y 2 .

    Setting them equal to zero yields the system of equations
    2 x 2 y 4 = 0 2 x + 8 y 2 = 0 .

    The solution to this system is x = 3 and y = 1 . Therefore ( 3 , 1 ) is a critical point of f . Calculating f ( 3 , 1 ) gives f ( 3 , 1 ) = 17 .
    The next step involves finding the extrema of f on the boundary of its domain. The boundary of its domain consists of four line segments as shown in the following graph:
    A rectangle is drawn in the first quadrant with one corner at the origin, horizontal length 4, and height 2. This rectangle is marked D, and the sides are marked in counterclockwise order from the side overlapping the x axis L1, L2, L3, and L4.
    Graph of the domain of the function f ( x , y ) = x 2 2 x y + 4 y 2 4 x 2 y + 24 .

    L 1 is the line segment connecting ( 0 , 0 ) and ( 4 , 0 ) , and it can be parameterized by the equations x ( t ) = t , y ( t ) = 0 for 0 t 4 . Define g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = t 2 4 t + 24 . Differentiating g leads to g ( t ) = 2 t 4 . Therefore, g has a critical value at t = 2 , which corresponds to the point ( 2 , 0 ) . Calculating f ( 2 , 0 ) gives the z- value 20 .
    L 2 is the line segment connecting ( 4 , 0 ) and ( 4 , 2 ) , and it can be parameterized by the equations x ( t ) = 4 , y ( t ) = t for 0 t 2 . Again, define g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = 4 t 2 10 t + 24 . Then, g ( t ) = 8 t 10 . g has a critical value at t = 5 4 , which corresponds to the point ( 0 , 5 4 ) . Calculating f ( 0 , 5 4 ) gives the z- value 27.75 .
    L 3 is the line segment connecting ( 0 , 2 ) and ( 4 , 2 ) , and it can be parameterized by the equations x ( t ) = t , y ( t ) = 2 for 0 t 4 . Again, define g ( t ) = f ( x ( t ) , y ( t ) ) . This gives g ( t ) = t 2 8 t + 36 . The critical value corresponds to the point ( 4 , 2 ) . So, calculating f ( 4 , 2 ) gives the z- value 20 .
    L 4 is the line segment connecting ( 0 , 0 ) and ( 0 , 2 ) , and it can be parameterized by the equations x ( t ) = 0 , y ( t ) = t for 0 t 2 . This time, g ( t ) = 4 t 2 2 t + 24 and the critical value t = 1 4 correspond to the point ( 0 , 1 4 ) . Calculating f ( 0 , 1 4 ) gives the z- value 23.75 .
    We also need to find the values of f ( x , y ) at the corners of its domain. These corners are located at ( 0 , 0 ) , ( 4 , 0 ) , ( 4 , 2 ) and ( 0 , 2 ) :
    f ( 0 , 0 ) = ( 0 ) 2 2 ( 0 ) ( 0 ) + 4 ( 0 ) 2 4 ( 0 ) 2 ( 0 ) + 24 = 24 f ( 4 , 0 ) = ( 4 ) 2 2 ( 4 ) ( 0 ) + 4 ( 0 ) 2 4 ( 4 ) 2 ( 0 ) + 24 = 24 f ( 4 , 2 ) = ( 4 ) 2 2 ( 4 ) ( 2 ) + 4 ( 2 ) 2 4 ( 4 ) 2 ( 2 ) + 24 = 20 f ( 0 , 2 ) = ( 0 ) 2 2 ( 0 ) ( 2 ) + 4 ( 2 ) 2 4 ( 0 ) 2 ( 2 ) + 24 = 36 .

    The absolute maximum value is 36 , which occurs at ( 0 , 2 ) , and the global minimum value is 20 , which occurs at both ( 4 , 2 ) and ( 2 , 0 ) as shown in the following figure.
    The function f(x, y) = x2 − 2xy – 4x + 4y2 – 2y + 24 is shown with local minima at (4, 2, 20) and (2, 0, 20) and local maximum at (0, 2, 36). The shape is a plane curving up on the corners significantly near (0, 2) and slightly less near (4, 0).
    The function f ( x , y ) has two global minima and one global maximum over its domain.
  2. Using the problem-solving strategy, step 1 involves finding the critical points of g on its domain. Therefore, we first calculate g x ( x , y ) and g y ( x , y ) , then set them each equal to zero:
    g x ( x , y ) = 2 x + 4 g y ( x , y ) = 2 y 6 .

    Setting them equal to zero yields the system of equations
    2 x + 4 = 0 2 y 6 = 0 .

    The solution to this system is x = −2 and y = 3 . Therefore, ( −2 , 3 ) is a critical point of g . Calculating g ( −2 , 3 ) , we get
    g ( −2 , 3 ) = ( −2 ) 2 + 3 2 + 4 ( −2 ) 6 ( 3 ) = 4 + 9 8 18 = −13 .

    The next step involves finding the extrema of g on the boundary of its domain. The boundary of its domain consists of a circle of radius 4 centered at the origin as shown in the following graph.
    A filled-in circle marked D of radius four with center at the origin.
    Graph of the domain of the function g ( x , y ) = x 2 + y 2 + 4 x 6 y .

    The boundary of the domain of g can be parameterized using the functions x ( t ) = 4 cos t , y ( t ) = 4 sin t for 0 t 2 π . Define h ( t ) = g ( x ( t ) , y ( t ) ) :
    h ( t ) = g ( x ( t ) , y ( t ) ) = ( 4 cos t ) 2 + ( 4 sin t ) 2 + 4 ( 4 cos t ) 6 ( 4 sin t ) = 16 cos 2 t + 16 sin 2 t + 16 cos t 24 sin t = 16 + 16 cos t 24 sin t .

    Setting h ( t ) = 0 leads to
    16 sin t 24 cos t = 0 16 sin t = 24 cos t −16 sin t −16 cos t = 24 cos t −16 cos t tan t = 4 3 .

    16 sin t 24 cos t = 0 16 sin t = 24 cos t −16 sin t −16 cos t = 24 cos t −16 cos t tan t = 3 2 .

    This equation has two solutions over the interval 0 t 2 π . One is t = π arctan ( 3 2 ) and the other is t = 2 π arctan ( 3 2 ) . For the first angle,
    sin t = sin ( π arctan ( 3 2 ) ) = sin ( arctan ( 3 2 ) ) = 3 13 13 cos t = cos ( π arctan ( 3 2 ) ) = cos ( arctan ( 3 2 ) ) = 2 13 13 .

    Therefore, x ( t ) = 4 cos t = 8 13 13 and y ( t ) = 4 sin t = 12 13 13 , so ( 8 13 13 , 12 13 13 ) is a critical point on the boundary and
    g ( 8 13 13 , 12 13 13 ) = ( 8 13 13 ) 2 + ( 12 13 13 ) 2 + 4 ( 8 13 13 ) 6 ( 12 13 13 ) = 144 13 + 64 13 32 13 13 72 13 13 = 208 104 13 13 −12.844 .

    For the second angle,
    sin t = sin ( 2 π arctan ( 3 2 ) ) = sin ( arctan ( 3 2 ) ) = 3 13 13 cos t = cos ( 2 π arctan ( 3 2 ) ) = cos ( arctan ( 3 2 ) ) = 2 13 13 .

    Therefore, x ( t ) = 4 cos t = 8 13 13 and y ( t ) = 4 sin t = 12 13 13 , so ( 8 13 13 , 12 13 13 ) is a critical point on the boundary and
    g ( 8 13 13 , 12 13 13 ) = ( 8 13 13 ) 2 + ( 12 13 13 ) 2 + 4 ( 8 13 13 ) 6 ( 12 13 13 ) = 144 13 + 64 13 + 32 13 13 + 72 13 13 = 208 + 104 13 13 44.844 .

    The absolute minimum of g is −13 , which is attained at the point ( −2 , 3 ) , which is an interior point of D . The absolute maximum of g is approximately equal to 44.844, which is attained at the boundary point ( 8 13 13 , 12 13 13 ) . These are the absolute extrema of g on D as shown in the following figure.
    The function f(x, y) = x2 + 4x + y2 – 6y is shown with local minimum at (–12/5, 16/5, –64/5) and local maximum at (12/5, −16/5, 224/5). The shape is a plane curving up from near (−4, 4) to (4, −4).
    The function f ( x , y ) has a local minimum and a local maximum.
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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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