Bounded, Closed, and Compact Sets

Transcription

1 Bounded, Closed, and Compact Sets Definition Let D be a subset of R n. Then D is said to be bounded if there is a number M > 0 such that x < M for all x D. D is closed if it contains all the boundary points. If D is both closed and bounded then it is said to be compact. Quang T. Bach Math 20C November 17, / 14

2 Question 1. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + y 2 + z 2 < 1} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, / 14

3 Question 2. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + z 2 < 1} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, / 14

4 Question 3. Consider the following subset of R 3 : U = {(x, y, z) : x 2 + y 2 + z 2 = 2} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, / 14

5 Question 4. Consider the following subset of R 3 : U = {(x, y, z) : x + y + z = 2} Then U is A. Neither closed nor bounded B. Closed but not bounded C. Bounded but not closed D. Compact, i.e. both closed and bounded Quang T. Bach Math 20C November 17, / 14

6 Lagrange s Multiplier Method for Finding Extrema Let f, g : R n R be C 1 functions. Suppose we want to find the maximum and minimum of f (x) restricted to the constraint g(x) = c, for some constant c. Theorem If x 0 is an extrema of f (x) subject to the constraint g(x) = c and if g(x 0 ) 0 then f (x 0 ) = λ g(x 0 ) for some scalar λ. Quang T. Bach Math 20C November 17, / 14

7 Some Remarks This theorem does not guarantee the existence of the extrema of f (x) subject to the constraint g(x) = c. It simply says that if there is an extrema, then here is how to find possible candidates for it. If there is no x 0 that satisfies f (x 0 ) = λ g(x 0 ) and g(x 0 ) = c, then f does not have an extreme subject to the constraint g(x) = c. The solution to the equation f (x 0 ) = λ g(x 0 ) is called the critical point of f subject to the constraint g(x) = c. Quang T. Bach Math 20C November 17, / 14

8 The Step Lagrange s Multiplier Method Let f, g : R n R be C 1 functions. Suppose we want to find the maximum and minimum of f (x) restricted to the constraint g(x) = c, for some constant c. We shall find the potential candidates for the extrema by solving the following system: f (x 1,..., x n ) = λ g (x 1,..., x n ) x 1 x 1 f (x 1,..., x n ) = λ g (x 1,..., x n ) x 2 x 2. f (x 1,..., x n ) = λ g (x 1,..., x n ) x n x n g(x 1,..., x n ) = c Quang T. Bach Math 20C November 17, / 14

9 Example Example Find the point on the circle x 2 + y 2 = 1 that is closest to the point (4, 2). The distance between any point (x, y) and (4, 2) is given by d(x, y) = (x 4) 2 + (y + 2) 2. Notice that d(x, y) is minimal if and only if (d(x, y)) 2 is minimal, so we shall consider the objective function subject to the constraint f (x, y) = (x 4) 2 + (y + 2) 2 g(x, y) = x 2 + y 2 = 1 Quang T. Bach Math 20C November 17, / 14

10 Example The Lagrange s multiplier conditions are: f (x, y) = λ g(x, y) g(x, y) = 1 2(x 4) = λ(2x) 2(y + 2) = λ(2y) x 2 + y 2 = 1 x 4 = xλ y + 2 = yλ x 2 + y 2 = 1 From the 1st eqn, we observe that x 0 which then leads to λ = x 4. x From the 2nd eqn, we observe that y 0 which then leads to λ = y + 2. y So x 4 = λ = y + 2 xy 4y = xy + 2x x = 2y. x y Now subs x = 2y to the last eqn: ( 2y) 2 + y 2 = 1 y = ±1/ 5. So x = 2y = 2/ 5. Quang T. Bach Math 20C November 17, / 14

11 Example We obtain two critical points ( 2 ) ( 1 2, and 5, 1 ) In this case, since the constraint is a closed and bounded set, f must have a maximum and minimum. In fact, these extrema are given by the solutions above. Here, ( f 2 ) 1, = f ( 2 5, 1 5 ) = ( 2 5, ) 1 is the maximum of f (x, y) on the unit circle and 5 So ( 2 5, 1 ) is the minimum of f on the unit circle. 5 Quang T. Bach Math 20C November 17, / 14

12 Example Example A manufacturer s production is modeled by the Cobb-Douglas function P(L, K) = 100L 3/4 K 1/4 where L represents the units of labor and K represents the units of capital. Each labor unit costs $200 and each capital unit costs $250. The total expenses for labor and capital is $50,000. Find the maximum production level. The objective function is subject to the constraint P(L, K) = 100L 3/4 K 1/4 g(l, K) = 200L + 250K = The Lagrange s multiplier conditions are: 75L 1/4 K 1/4 = 200λ Quang T. Bach Math 20C November 17, / 14

13 Example From the first two equations we obtain 3 8 L 1/4 K 1/4 = λ = 1 10 L3/4 K 3/ K = L. ( ) 15 Now from the last equation: K + 250K = K = 50. So L = Quang T. Bach Math 20C November 17, / 14

14 Example Here, we obtain a critical point (187.5, 50). The constraint 200L + 250K = represents a line segment which is closed and bounded so P(L, K) must have a max and a min. The extrema must occur at either the critical point or the end points. P(0, 200) = P(250, 0) = 0 P(187.5, 50) so (187.5, 50) is the maximum of P(L, K) subject to the constraint 200L + 250K = The maximum production level is Quang T. Bach Math 20C November 17, / 14

15 Example - HW6 Problem 4a Example Write the number 120 as a sum of three numbers so that the sum of the products between any pair of two numbers is a maximum. Solution: Let x, y, z denote the three numbers. The objective function is subject to the constraint f (x, y, z) = xy + xz + yz g(x, y, z) = x + y + z = 120. Quang T. Bach Math 20C November 17, / 14

16 Example - HW6 Problem 4a The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 120 y + z = λ x + z = λ x + y = λ x + y + z = 120 The first three equations give x = y = z. Substitute this into the fourth equation to obtain x = y = z = 40. So (x, y, z) = (40, 40, 40) is the only critical point obtained from the method. Here, f (40, 40, 40) = Quang T. Bach Math 20C November 17, / 14

17 Example - HW6 Problem 4b Example Find three positive numbers x, y, and z whose sum is 10 such that x 2 y 2 z is a maximum. Solution: Let 0 < x, y, z < 10 denote the three numbers. The objective function is f (x, y, z) = x 2 y 2 z subject to the constraint g(x, y, z) = x + y + z = 10. Quang T. Bach Math 20C November 17, / 14

18 Example - HW6 Problem 4b The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 10 2xy 2 z = λ 2x 2 yz = λ x 2 y 2 = λ x + y + z = 10 The first three equations give x = y = 2z. Substitute this into the fourth equation to obtain x = y = 4 and z = 2. So (x, y, z) = (4, 4, 2) is the only critical point obtained from the method. Here, f (4, 4, 2) = 512. Quang T. Bach Math 20C November 17, / 14

19 Example - HW6 Problem 4c Example A wooden box is to be made with $ worth of wood. The lid is to be made from wood that costs $2.00 per square inch and the rest of the box is to be made from wood that costs $2.50 per square inch. What is the biggest box that could be made? Solution: Let x, y, z > 0 denote the length, width, and height of the box. The objective function is subject to the constraint f (x, y, z) = xyz g(x, y, z) = 4.5xy + 5xz + 5yz = 120. Quang T. Bach Math 20C November 17, / 14

20 Example - HW6 Problem 4c The Lagrange s multiplier conditions are: f (x, y, z) = λ g(x, y, z) g(x, y, z) = 120 xz = λ(4.5x + 5z) yz = λ(4.5y + 5z) xy = λ(5x + 5y) 4.5xy + 5yz + 5xz = 120 Recall that x, y, z > 0 so from the first two equations, we obtain: xz 4.5x + 5z = λ = yz z x = y. With x = y, now from the third equation: x 2 = 10λx λ = x/10. Now subs λ = x/10 back to the first equation to obtain: xz = x 9x (4.5x + 5z) z = Quang T. Bach Math 20C November 17, / 14

21 Example - HW6 Problem 4c Finally, we subs x = y, z = 9x/10 into the last equation to obtain: 4.5xy + 5yz + 5xz = x 2 = 120 x 2 = 80 9 x = The solution ( obtain from the Lagrange s multiplier method is 4 5 (x, y, z) = 3, 4 5 3, 6 ) ( and f 5 3, 4 5 3, 6 ) 5 = Quang T. Bach Math 20C November 17, / 14

22 Some Final Remarks In all 3 examples from HW6 above, I never claimed that the solution obtained from the Lagrange s multiplier method is the maximum of f (x, y, z) subject to the constraint g(x, y, z) = c. There is no guarantee that the solutions obtained from Lagrange s multiplier method is the extrema of the function. In fact, in all three cases above, the constraint region is not closed and bounded, so there is no guarantee that a global max/min even exists! Even thought the solutions we obtained here are all maximum of f, further analysis is required to obtain this conclusion. Quang T. Bach Math 20C November 17, / 14