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Use the method of Lagrange multipliers to find the minimum value of the function

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

subject to the constraints 2 x + y + 2 z = 9 and 5 x + 5 y + 7 z = 29 .

f ( 2 , 1 , 2 ) = 9 is a minimum.

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Key concepts

  • An objective function combined with one or more constraints is an example of an optimization problem.
  • To solve optimization problems, we apply the method of Lagrange multipliers using a four-step problem-solving strategy.

Key equations

  • Method of Lagrange multipliers, one constraint
    f ( x 0 , y 0 ) = λ g ( x 0 , y 0 ) g ( x 0 , y 0 ) = 0
  • Method of Lagrange multipliers, two constraints
    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

For the following exercises, use the method of Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.

f ( x , y ) = x 2 y ; x 2 + 2 y 2 = 6

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f ( x , y , z ) = x y z , x 2 + 2 y 2 + 3 z 2 = 6

maximum: 2 3 3 , minimum: −2 3 3

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f ( x , y ) = x y ; 4 x 2 + 8 y 2 = 16

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f ( x , y ) = 4 x 3 + y 2 ; 2 x 2 + y 2 = 1

maximum: ( 2 2 , 0 , 2 ) , minimum: ( 2 2 , 0 , 2 )

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f ( x , y , z ) = x 2 + y 2 + z 2 , x 4 + y 4 + z 4 = 1

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f ( x , y , z ) = y z + x y , x y = 1 , y 2 + z 2 = 1

maximum: 3 2 , minimum = 1

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f ( x , y ) = x 2 + y 2 , ( x 1 ) 2 + 4 y 2 = 4

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f ( x , y ) = 4 x y , x 2 9 + y 2 16 = 1

maxima: f ( 3 2 2 , 2 2 ) = 24 , f ( 3 2 2 , −2 2 ) = 24 ; minima: f ( 3 2 2 , 2 2 ) = −24 , f ( 3 2 2 , −2 2 ) = −24

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f ( x , y , z ) = x + y + z , 1 x + 1 y + 1 z = 1

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f ( x , y , z ) = x + 3 y z , x 2 + y 2 + z 2 = 4

maximum: 2 11 at f ( 2 11 , 6 11 , −2 11 ) ; minimum: −2 11 at f ( −2 11 , −6 11 , 2 11 )

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f ( x , y , z ) = x 2 + y 2 + z 2 , x y z = 4

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Minimize f ( x , y ) = x 2 + y 2 on the hyperbola x y = 1 .

2.0

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Minimize f ( x , y ) = x y on the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 .

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Maximize f ( x , y , z ) = 2 x + 3 y + 5 z on the sphere x 2 + y 2 + z 2 = 19 .

19 2

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Maximize f ( x , y ) = x 2 y 2 ; x > 0 , y > 0 ; g ( x , y ) = y x 2 = 0

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The curve x 3 y 3 = 1 is asymptotic to the line y = x . Find the point(s) on the curve x 3 y 3 = 1 farthest from the line y = x .

( 1 2 3 , −1 2 3 )

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Maximize U ( x , y ) = 8 x 4 / 5 y 1 / 5 ; 4 x + 2 y = 12

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Minimize f ( x , y ) = x 2 + y 2 , x + 2 y 5 = 0 .

f ( 1 , 2 ) = 5

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Maximize f ( x , y ) = 6 x 2 y 2 , x + y 2 = 0 .

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Minimize f ( x , y , z ) = x 2 + y 2 + z 2 , x + y + z = 1 .

f ( 1 3 , 1 3 , 1 3 ) = 1 3

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Minimize f ( x , y ) = x 2 y 2 subject to the constraint x 2 y + 6 = 0 .

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Minimize f ( x , y , z ) = x 2 + y 2 + z 2 when x + y + z = 9 and x + 2 y + 3 z = 20 .

minimum: f ( 2 , 3 , 4 ) = 29

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For the next group of exercises, use the method of Lagrange multipliers to solve the following applied problems.

A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the diagram. If the perimeter of the pentagon is 10 in., find the lengths of the sides of the pentagon that will maximize the area of the pentagon.

A rectangle with an isosceles triangle on top. The side of the isosceles triangle with the two equal angles of size θ overlaps the top length of the rectangle.
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A rectangular box without a top (a topless box) is to be made from 12 ft 2 of cardboard. Find the maximum volume of such a box.

The maximum volume is 4 ft 3 . The dimensions are 1 × 2 × 2 ft.

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Find the minimum and maximum distances between the ellipse x 2 + x y + 2 y 2 = 1 and the origin.

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Find the point on the surface x 2 2 x y + y 2 x + y = 0 closest to the point ( 1 , 2 , −3 ) .

( 1 , 1 2 , −3 )

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Show that, of all the triangles inscribed in a circle of radius R (see diagram), the equilateral triangle has the largest perimeter.

A circle with an equilateral triangle drawn inside of it such that each vertex of the triangle touches the circle.
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Find the minimum distance from point ( 0 , 1 ) to the parabola x 2 = 4 y .

1.0

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Find the minimum distance from the parabola y = x 2 to point ( 0 , 3 ) .

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Find the minimum distance from the plane x + y + z = 1 to point ( 2 , 1 , 1 ) .

3

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A large container in the shape of a rectangular solid must have a volume of 480 m 3 . The bottom of the container costs $5/m 2 to construct whereas the top and sides cost $3/m 2 to construct. Use Lagrange multipliers to find the dimensions of the container of this size that has the minimum cost.

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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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