MATH 19520/51 Class 10

Transcription

1 MATH 19520/51 Class 10 Minh-Tam Trinh University of Chicago

2 1 Method of Lagrange multipliers. 2 Examples of Lagrange multipliers.

3 The Problem The ingredients: 1 A set of parameters, say x 1,..., x n. 2 A function of those parameters, say f (x 1,..., x n ), that we want to optimize. 3 A constraint g(x 1,..., x n ) = c on the parameters. The problem: Optimize the function subject to the constraint.

4 Rephrased: Find the global maxima/minima of f relative to a level set of g.

5 Rephrased: Find the global maxima/minima of f relative to a level set of g. (Above, the red curve is not the level curve, but the value of f along the level curve.)

6 Example Suppose we need to build the rectangular garden with the largest possible area, given exactly 100 m of fencing.

7 Example Suppose we need to build the rectangular garden with the largest possible area, given exactly 100 m of fencing. 1 The parameters are the dimensions of the garden, say length l and width w.

8 Example Suppose we need to build the rectangular garden with the largest possible area, given exactly 100 m of fencing. 1 The parameters are the dimensions of the garden, say length l and width w. 2 The function to maximize is the area A = l w.

9 Example Suppose we need to build the rectangular garden with the largest possible area, given exactly 100 m of fencing. 1 The parameters are the dimensions of the garden, say length l and width w. 2 The function to maximize is the area A = l w. 3 The constraint is that its perimeter P = 2l + 2w must equal 100 m, i.e., P = 100 m. Technically, another constraint is that only l 0 and w 0 make sense.

10 Example Suppose we want to find the point(s) on the level surface (x + y + z) 2 = 9 closest to the origin.

11 Example Suppose we want to find the point(s) on the level surface (x + y + z) 2 = 9 closest to the origin. 1 The parameters are the coordinates x, y, z of the point.

12 Example Suppose we want to find the point(s) on the level surface (x + y + z) 2 = 9 closest to the origin. 1 The parameters are the coordinates x, y, z of the point. 2 The function to minimize is its distance r from (0, 0, 0), i.e., r(x, y, z) = x 2 + y 2 + z 2. Actually, it s easier to work with R(x, y, z) = r(x, y, z) 2.

13 Example Suppose we want to find the point(s) on the level surface (x + y + z) 2 = 9 closest to the origin. 1 The parameters are the coordinates x, y, z of the point. 2 The function to minimize is its distance r from (0, 0, 0), i.e., r(x, y, z) = x 2 + y 2 + z 2. Actually, it s easier to work with R(x, y, z) = r(x, y, z) 2. 3 The constraint is that the point belongs to the surface, i.e., g(x, y, z) = 1 where g(x, y, z) = (x + y + z) 2.

14 The Method We want the global maxima/minima of f relative to the level set g = c.

15 The Method We want the global maxima/minima of f relative to the level set g = c. We need to find and compare the values of f at: 1 Its local maxima/minima relative to the level set. 2 Boundary points of the level set. Lagrange saw how to identify the relative local maxima/minima.

16 When f attains local maxima/minima relative to g = c, the level curves of f and g are tangent. Thus f and are parallel.

17 When f attains local maxima/minima relative to g = c, the level curves of f and g are tangent. Thus f and are parallel.

18 Lagrange s algorithm, stated for two variables:

19 Lagrange s algorithm, stated for two variables: 1 Find all points on the level curve g(x, y) = c where f (x, y) and g(x, y) are parallel, i.e., (1) f (x, y) = λ g(x, y) for some scalar λ, the Lagrange multiplier at the point. These points are the local extrema of f relative to the level curve.

20 Lagrange s algorithm, stated for two variables: 1 Find all points on the level curve g(x, y) = c where f (x, y) and g(x, y) are parallel, i.e., (1) f (x, y) = λ g(x, y) for some scalar λ, the Lagrange multiplier at the point. These points are the local extrema of f relative to the level curve. 2 Find all points on the level curve that belong to the boundary of the domain of g.

21 Lagrange s algorithm, stated for two variables: 1 Find all points on the level curve g(x, y) = c where f (x, y) and g(x, y) are parallel, i.e., (1) f (x, y) = λ g(x, y) for some scalar λ, the Lagrange multiplier at the point. These points are the local extrema of f relative to the level curve. 2 Find all points on the level curve that belong to the boundary of the domain of g. 3 Compare the values of f at all the points from (1) and (2) to find the global extrema of f relative to the level curve.

22 Examples Example In our garden problem, we want to maximize A = l w subject to P = 2l + 2w = 100 m.

23 Examples Example In our garden problem, we want to maximize A = l w subject to P = 2l + 2w = 100 m. 1 A = (A l, A w ) = (w,l ) and P = (P l, P w ) = (2, 2), so the identity A = λ P becomes (2) (w,l ) = (2λ, 2λ). On the level curve 2l + 2w = 100 m, this only occurs at (l, w) = (25 m, 25 m), where λ = 25/2 m.

24 Examples Example In our garden problem, we want to maximize A = l w subject to P = 2l + 2w = 100 m. 1 A = (A l, A w ) = (w,l ) and P = (P l, P w ) = (2, 2), so the identity A = λ P becomes (2) (w,l ) = (2λ, 2λ). On the level curve 2l + 2w = 100 m, this only occurs at (l, w) = (25 m, 25 m), where λ = 25/2 m. 2 The domain is {(l, w) : l, w 0}, so the boundary points of the level curve are (l, w) = (0 m, 50 m) and (50 m, 0 m).

25 Examples Example In our garden problem, we want to maximize A = l w subject to P = 2l + 2w = 100 m. 1 A = (A l, A w ) = (w,l ) and P = (P l, P w ) = (2, 2), so the identity A = λ P becomes (2) (w,l ) = (2λ, 2λ). On the level curve 2l + 2w = 100 m, this only occurs at (l, w) = (25 m, 25 m), where λ = 25/2 m. 2 The domain is {(l, w) : l, w 0}, so the boundary points of the level curve are (l, w) = (0 m, 50 m) and (50 m, 0 m). 3 The maximum occurs at (l, w) = (25 m, 25 m), which gives us A = 625 m 2.

26 Example In our distance problem, we must minimize R = x 2 + y 2 + z 2 subject to g = 9, where g = (x + y + z) 2.

27 Example In our distance problem, we must minimize R = x 2 + y 2 + z 2 subject to g = 9, where g = (x + y + z) 2. 1 R = (2x, 2y, 2z) and g = (2(x + y + z), 2(x + y + z), 2(x + y + z)), so we seek (3) (2x, 2y, 2z) = (2λ(x + y + z), 2λ(x + y + z), 2λ(x + y + z)). Then x = y = z, so the relative extrema on the level surface are (x, y, z) = (1, 1, 1) and ( 1, 1, 1), with λ = 1/3.

28 Example In our distance problem, we must minimize R = x 2 + y 2 + z 2 subject to g = 9, where g = (x + y + z) 2. 1 R = (2x, 2y, 2z) and g = (2(x + y + z), 2(x + y + z), 2(x + y + z)), so we seek (3) (2x, 2y, 2z) = (2λ(x + y + z), 2λ(x + y + z), 2λ(x + y + z)). Then x = y = z, so the relative extrema on the level surface are (x, y, z) = (1, 1, 1) and ( 1, 1, 1), with λ = 1/3. 2 The domain of g is all of R 3, so no boundary points.

29 Example In our distance problem, we must minimize R = x 2 + y 2 + z 2 subject to g = 9, where g = (x + y + z) 2. 1 R = (2x, 2y, 2z) and g = (2(x + y + z), 2(x + y + z), 2(x + y + z)), so we seek (3) (2x, 2y, 2z) = (2λ(x + y + z), 2λ(x + y + z), 2λ(x + y + z)). Then x = y = z, so the relative extrema on the level surface are (x, y, z) = (1, 1, 1) and ( 1, 1, 1), with λ = 1/3. 2 The domain of g is all of R 3, so no boundary points. 3 Both (1, 1, 1) and ( 1, 1, 1) give R = 3.

30 Variations Variation 1: Constraints given by inequalities, not equations. Example How to optimize f (x, y) subject to the constraints a g(x, y) b?

31 Variations Variation 1: Constraints given by inequalities, not equations. Example How to optimize f (x, y) subject to the constraints a g(x, y) b? Each value a c b gives us a level curve g(x, y) = c within this region. Run the Lagrange-multiplier step for each level curve individually.

32 Variations Variation 1: Constraints given by inequalities, not equations. Example How to optimize f (x, y) subject to the constraints a g(x, y) b? Each value a c b gives us a level curve g(x, y) = c within this region. Run the Lagrange-multiplier step for each level curve individually. To find the global optima on the whole region, compare the points from all of the level curves together.

33 Variation 2: Two constraint equations instead of one. Example How to optimize f (x, y, z) subject to g 1 (x, y, z) = c 1 and g 2 (x, y, z) = c 2?

34 Variation 2: Two constraint equations instead of one. Example How to optimize f (x, y, z) subject to g 1 (x, y, z) = c 1 and g 2 (x, y, z) = c 2? f is constrained to the intersection of two level sets.

35 Variation 2: Two constraint equations instead of one. Example How to optimize f (x, y, z) subject to g 1 (x, y, z) = c 1 and g 2 (x, y, z) = c 2? f is constrained to the intersection of two level sets. f is a linear combination of g 1 and g 2 at its local extrema relative to the intersection. Thus, two Lagrange multipliers: (4) f (x, y, z) = λ 1 g 1 (x, y, z) + λ 2 g 2 (x, y, z). Solve for (x, y, z) satisfying both g 1 (x, y, z) = c 1 and g 2 (x, y, z) = c 2.