6.1 Matrices. Definition: A Matrix A is a rectangular array of the form. A 11 A 12 A 1n A 21. A 2n. A m1 A m2 A mn A 22.

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1 61 Matrices Definition: A Matrix A is a rectangular array of the form A 11 A 12 A 1n A 21 A 22 A 2n A m1 A m2 A mn The size of A is m n, where m is the number of rows and n is the number of columns The entry A ij locates in the ith row and the jth column Lecture Notes for Math 1540 First Previous Next Last 1

2 61 Matrices Equality of two matrices: A = B if and only if 1 A and B have the same size 2 The corresponding entries are equal: A ij = B ij Transpose of a matrix: 1 If A has the size m n, then A T has the size n m 2 The ith row of A T is the ith column of A 3 (A T ) T = A Special matrices: 1 zero matrix 2 square matrix: diagonal matrix; upper/lower triangular matrix Lecture Notes for Math 1540 First Previous Next Last 2

3 62 Matrix Addition and Scalar Multiplication Matrix addition: (A + B) ij = A ij + B ij 1 commutative property: A + B = B + A 2 associative property: A + (B + C) = (A + B) + C 3 identity property: A + 0 = 0 + A = A Scalar multiplication: (ka) ij = ka ij Some properties: 1 k(a + B) = ka + kb 2 (k + l)a = ka + la 3 k(la) = (kl)a 4 0A = 0 and k0 = 0 5 (A + B) T = A T + B T and (ka) T = ka T Lecture Notes for Math 1540 First Previous Next Last 3

4 63 Matrix Multiplication Definition Let A be an m n matrix, and B be an n p matrix Then 1 AB is an m p matrix 2 (AB) ik = n A ij B jk = A i1 B 1k + A i2 B 2k + + A in B nk j=1 The number of columns of A should be equal to the number of rows of B (AB) ik is the product of ith row of A and kth column of B Matrix addition and matrix multiplication 1 A has size m n and B has size m n, then A + B has size m n 2 A has size m n and B has size n p, then AB has size m p Lecture Notes for Math 1540 First Previous Next Last 4

5 63 Matrix Multiplication Properties: 1 associative: A(BC) = (AB)C 2 distributive: A(B + C) = AB + AC and (A + B)C = AC + BC 3 k(ab) = (ka)b = A(kB) 4 (AB) T = B T A T 5 In general, AB BA and (AB) T A T B T (Matrix multiplication is non-commutative!) If A is a square matrix of order n, I is an identity matrix of order n and 0 is a zero matrix of size n n, then AI = IA = A and A0 = 0A = 0 Power of a square matrix: A 2 = A A, A 3 = A A A, Lecture Notes for Math 1540 First Previous Next Last 5

6 63 Matrix Multiplication System of linear equations A 11 X 1 + A 12 X A 1n X n = B 1 A 21 X 1 + A 22 X A 2n X n = B 2 A m1 X 1 + A m2 X A mn X n = B m Matrix equation AX = B 1 Coefficient matrix A of size m n is known 2 The m 1 column vector B is also given 3 The n 1 column vector X is the variable to be solved 4 Solution C is a column vector of size n 1 such that AC = B Lecture Notes for Math 1540 First Previous Next Last 6

7 64 Solving Systems by Reducing Matrices Three elementary row operations: 1 R i R j : interchange rows R i and R j 2 kr i : multiply row R i by a nonzero constant k 3 kr i + R j : add k times row R i to row R j (but leave R i unchanged) Reduced matrix: 1 All zero-rows are at the bottom of the matrix 2 The leading entry in each nonzero-row is 1, and all others entries in the column of the leading entry is 0 3 The leading entry in each nonzero-row is to the right of the leading entry in any row above it Lecture Notes for Math 1540 First Previous Next Last 7

8 65 Solving Systems by Reducing Matrices (continued) Homogeneous system: AX = 0, or equivalently, A 11 X 1 + A 12 X A 1n X n = 0 A 21 X 1 + A 22 X A 2n X n = 0 A m1 X 1 + A m2 X A mn X n = 0 X = 0 is always a trivial solution To solve a homogeneous system, we only need to reduce the coefficient matrix A, instead of the augmented coefficient matrix [A 0] Lecture Notes for Math 1540 First Previous Next Last 8

9 65 Solving Systems by Reducing Matrices (continued) Homogeneous system: AX = 0, or equivalently, A 11 X 1 + A 12 X A 1n X n = 0 A 21 X 1 + A 22 X A 2n X n = 0 A m1 X 1 + A m2 X A mn X n = 0 The number of equations m is equal to the number of rows of A The number of unknowns n is equal to the number of columns of A If A is a reduced matrix with k nonzero-rows, then k m Lecture Notes for Math 1540 First Previous Next Last 9

10 65 Solving Systems by Reducing Matrices (continued) Theorem Let A be a reduced coefficient matrix of size m n for a homogeneous system AX = 0 (Note that X is an n 1 column vector and 0 is an m 1 column vector) 1 If A has exactly k nonzero-rows, then k n 2 If A has exactly k nonzero-rows and k < n, then the system has infinitely many solutions 3 If A has exactly k nonzero-rows and k = n, then the system has a unique solution X = 0 (the trivial solution) Corollary If m < n, (the number of equations is less than the number of unknowns), then k m < n, which by the above theorem implies that the homogeneous system has infinitely many solutions Lecture Notes for Math 1540 First Previous Next Last 10

11 66 Inverses A square matrix A of order n is invertible if there exists a square matrix C of order n such that CA = AC = I, where I is the identity matrix of order n We use A 1 to denote the inverse of A A should be a square matrix Recall that AI = IA = A for any square matrix A having the same order of I The identity matrix I is invertible and I 1 = I If A is invertible, then A 1 is unique and AA 1 = A 1 A = I, namely, multiplication of A and A 1 is commutative Recall that in general, matrix multiplication is NOT commutative Lecture Notes for Math 1540 First Previous Next Last 11

12 66 Inverses Find the inverse of A = [ ] Let C be the inverse of A such that AC = I, where [ ] x z C = y w It follows that [ ] [ ] 1 2 x z 3 7 y w = [ x + 2y ] z + 2w 3x + 7y 3z + 7w = [ ] Lecture Notes for Math 1540 First Previous Next Last 12

13 66 Inverses We have two linear systems { x + 2y = 1 3x + 7y = 0 and { z + 2w = 0 3z + 7w = 1 Reducing augmented coefficient matrix [ ] 1 0 3R 1 +R 2 [ ] 1 3 2R 2 +R 1 [ ] 7 3 { x = 7 y = 3 [ ] 0 1 3R 1 +R 2 [ ] 0 1 2R 2 +R 1 [ ] 2 1 { z = 2 w = 1 Lecture Notes for Math 1540 First Previous Next Last 13

14 66 Inverses Reducing augmented coefficient matrix [ ] 1 0 3R 1 +R 2 [ ] 1 3 2R 2 +R 1 [ ] 7 3 { x = 7 y = 3 [ ] 0 1 3R 1 +R 2 [ ] 0 1 2R 2 +R 1 [ ] 2 1 { z = 2 w = 1 Combining the above two reducing procedures [ ] R 1 +R 2 [ ] R 2 +R 1 [ ] Lecture Notes for Math 1540 First Previous Next Last 14

15 66 Inverses Find the inverse of A = [ ] Reducing the matrix [A I] to [I C] [ ] [ ] The inverse matrix A 1 = C = [ 7 ] Lecture Notes for Math 1540 First Previous Next Last 15

16 66 Inverses If A is square and [A I] can be reduced to the form [I C], then A is invertible and A 1 = C If A is square and [A I] is reduced to [R C] but R I, then A is not invertible If A is not square then A is not invertible If the square matrix A is invertible, then the matrix equation AX = B has a unique solution X = A 1 B If the coefficient matrix A is square but not invertible, then the homogeneous equation AX = 0 has infinitely many solutions Lecture Notes for Math 1540 First Previous Next Last 16

17 66 Inverses Matrix Algebra Algebra A is a square matrix of order n a is a real number I is the identity matrix of order n 1 is the real identity AI = IA = A a 1 = 1 a = a I 1 = I 1 1 = 1 If A is invertible, then A 1 is unique If a is invertible, then a 1 is unique AA 1 = A 1 A = I aa 1 = a 1 a = 1 If A 0, A may still not be invertible If a 0, then a is invertible In general AB BA ab = ba Lecture Notes for Math 1540 First Previous Next Last 17

18 67 Leontief s Input-Output Analysis Input-output matrix Input Sector 1 Sector 2 External Demand Output Totals Sector Sector Other Totals [ ] Leontief matrix A = Lecture Notes for Math 1540 First Previous Next Last 18

19 67 Leontief s Input-Output Analysis Leontief matrix A = To produce one unit of sector 1, it spends To produce one unit of sector 2, it spends on sector 1, and on sector 1, and on sector on sector Lecture Notes for Math 1540 First Previous Next Last 19

20 67 Leontief s Input-Output Analysis Leontief matrix A = External Demand matrix Production matrix D = X = [ ] [ ] Lecture Notes for Math 1540 First Previous Next Last 20

21 67 Leontief s Input-Output Analysis Production=Internal Demand+External Demand: X = AX + D 1200 = Solving production matrix X from external demand matrix D: X = AX + D X AX = D (I A)X = D X = (I A) 1 D Lecture Notes for Math 1540 First Previous Next Last 21

22 67 Leontief s Input-Output Analysis 1 Determine the Leontief matrix A 2 Find the external demand matrix D 3 Solve the production matrix X by X = (I A) 1 D Lecture Notes for Math 1540 First Previous Next Last 22

23 67 Leontief s Input-Output Analysis A very simple economy consists of two sectors: agriculture and milling To produce one unit of agricultural products requires 1 3 of a unit of agricultural products and 1 4 of a unit of milled products To produce one unit of milled products requires 3 4 of a unit of agricultural products and no units of milled products Determine the production levels needed to satisfy an external demand for 300 units of agriculture and 500 units of milled products Lecture Notes for Math 1540 First Previous Next Last 23

24 67 Leontief s Input-Output Analysis A very simple economy consists of two sectors: agriculture and milling To produce one unit of agricultural products requires 1 3 of a unit of agricultural products and 1 4 of a unit of milled products To produce one unit of milled products requires 3 4 of a unit of agricultural products and no units of milled products Determine the production levels needed to satisfy an external demand for 300 units of agriculture and 500 units of milled products 1 Leontief matrix 2 External demand matrix A = D = 3 Production matrix X = (I A) 1 D [ ] Lecture Notes for Math 1540 First Previous Next Last 24

25 67 Leontief s Input-Output Analysis Find the inverse of I A: R R R 1+R R 2+R Production matrix X = (I A) 1 D = = Lecture Notes for Math 1540 First Previous Next Last 25

26 67 Leontief s Input-Output Analysis Input-output matrix Production matrix X = [ ] = [ ] Leontief matrix A = = Lecture Notes for Math 1540 First Previous Next Last 26

27 67 Leontief s Input-Output Analysis Leontief matrix A = = Question: determine the corresponding production matrix if the external demand matrix changes to D = Lecture Notes for Math 1540 First Previous Next Last 27

28 67 Leontief s Input-Output Analysis Find the inverse of I A: R R R 1+R R 2+R Production matrix X = (I A) 1 D = = Lecture Notes for Math 1540 First Previous Next Last 28

29 71 Linear Inequalities in Two Variables Linear inequality in variables x and y (either a or b is nonzero): ax + by + c < 0 ax + by + c 0 ax + by + c > 0 ax + by + c 0 The nonvertical line y = mx + b separates the plane into three distinct parts: 1 the line itself, y = mx + b 2 the region (open half-plane) above the line, y > mx + b 3 the region (open half-plane) below the line, y < mx + b The vertical line x = a separates the plane into three distinct parts: 1 the line itself, x = a 2 the region (open half-plane) to the left of the line, x < a 3 the region (open half-plane) to the right of the line, x > a Lecture Notes for Math 1540 First Previous Next Last 29

30 72 Linear Programming If the feasible region is nonempty, closed and bounded, then the objective function has a maximum (minimum) value which is achieved at a corner point If the feasible region is empty, then the objective function has no maximum (minimum) Suppose the feasible region is nonempty, closed and unbounded 1 If there exists an isovalue (isocost) line below the feasible region, then the objective function has a minimum which is achieved at a corner point Otherwise, the objective function has no minimum 2 If there exists an isovalue (isoprofit) line above the feasible region, then the objective function has a maximum which is achieved at a corner point Otherwise, the objective function has no maximum Lecture Notes for Math 1540 First Previous Next Last 30

31 73 Multiple Optimum Solutions If (x 1, y 1 ) and (x 2, y 2 ) are two corner points at which an objective function is optimum, then the objective function will also be optimum at all points (x, y), where x = (1 t)x 1 + tx 2 y = (1 t)y 1 + ty 2 with 0 t 1 The above two expressions can be also written in terms of matrix form: [ ] [ ] [ ] x x1 x2 = (1 t) + t y y 1 y 2 Lecture Notes for Math 1540 First Previous Next Last 31

32 74 The Simplex Method Standard Linear Programming Problem: Maximize the linear function Z = c 1 x 1 + c 2 x c n x n subject to a 11 x 1 + a 12 x a 1n x n b 1 a 21 x 1 + a 22 x a 2n x n b 2 a m1 x 1 + a m2 x a mn x n b m where x 1, x 2,, x n and b 1, b 2,, b m are nonnegative Lecture Notes for Math 1540 First Previous Next Last 32

33 74 The Simplex Method Add slack variables s 1, s 2,, s m : a 11 x 1 + a 12 x a 1n x n + s 1 = b 1 a 21 x 1 + a 22 x a 2n x n + s 2 = b 2 a m1 x 1 + a m2 x a mn x n + s m = b m where x 1, x 2,, x n, s 1, s 2,, s m, and b 1, b 2,, b m are nonnegative From Z = c 1 x 1 + c 2 x c n x n we have c 1 x 1 c 2 x 2 c n x n + Z = 0 Lecture Notes for Math 1540 First Previous Next Last 33

34 74 The Simplex Method Initial Simplex Table: s 1 s 2 s m Z x 1 x 2 x n s 1 s 2 s m Z a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn c 1 c 2 c n Basic feasible solution (BFS): 1 Basic variables: s 1 = b 1, s 2 = b 2,, s m = b m 2 Nonbasic variables: x 1 = x 2 = = x n = 0 3 Objective function: Z = 0 b 1 b 2 b m 0 Lecture Notes for Math 1540 First Previous Next Last 34

35 74 The Simplex Method Each basic feasible solution (BFS) corresponds to a corner point in feasible region The simplex method provides an algorithm to search the BFS (corner point) where the objective function achieves its maximum Two principles of simplex method: 1 Eliminate negative indicators (start from the most negative indicator first) 2 Keep the last column nonnegative (normalize the row with smallest quotient) Lecture Notes for Math 1540 First Previous Next Last 35

36 75 Degeneracy, Unbounded Solutions, and Multiple Solutions Degeneracy occurs when there are two rows have the smallest quotient, or when the smallest quotient is zero If there exists no quotient, then the linear programming problem has an unbounded solution If the indicators are all nonnegative and the number of zero indicators is larger than the number of basic variables, then the linear programming problem has multiple solutions Lecture Notes for Math 1540 First Previous Next Last 36

37 76 Artificial Variables & 77 Minimization We need to add an artificial variable t when the feasible region does not contain the origin (ie, 0 is not a solution to the constraints) The objective function becomes W = Z Mt = c 1 x 1 + c 2 x c n x n Mt with M being a large parameter 1 The case a i1 x 1 + a i2 x a in x n b i with b i > 0 We add a slack variable s i 0 and an artificial variable t such that a i1 x 1 + a i2 x a in x n s i + t = b i 2 The case a i1 x 1 + a i2 x a in x n = b i with b i > 0 Since x 1 = 0, x 2 = 0,, x n = 0 is not a solution to this constraint, we need to add an artificial variable t such that a i1 x 1 + a i2 x a in x n + t = b i Minimizing the objective function Z = c 1 x 1 + c 2 x c n x n is the same as maximizing the objective function Z = c 1 x 1 c 2 x 2 c n x n Lecture Notes for Math 1540 First Previous Next Last 37

38 78 The Dual Maximize Minimize Z = c 1 x 1 + c 2 x c n x n W = b 1 y 2 + b 2 y b m y m subject to subject to a 11 x 1 + a 12 x a 1n x n b 1 a 11 y 1 + a 21 y a m1 y m c 1 a 21 x 1 + a 22 x a 2n x n b 2 a 12 y 1 + a 22 y a m2 y m c 2 a m1 x 1 + a m2 x a mn x n b m a 1n y 1 + a 2n y a mn y m c n and and x 1, x 2,, x n 0 y 1, y 2,, y m 0 Lecture Notes for Math 1540 First Previous Next Last 38

39 78 The Dual Maximize x 1 x n Minimize Z = [c 1, c 2,, c n ] x 2 W = [b 1, b 2,, b m ] subject to a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn x 1 x n x 2 0 x 1 x 2 x n b 1 b 2 b m subject to a 11 a 21 a m1 a 12 a 22 a m2 a 1n a 2n a mn y 1 y 2 0 y m y 1 y 2 y m y 1 y 2 y m c 1 c 2 c n Lecture Notes for Math 1540 First Previous Next Last 39

40 78 The Dual Maximize Z = CX Minimize W = B T Y subject to subject to AX B and X 0 A T Y C T and Y 0 where C = [c 1, c 2,, c n ], X = [x 1, x 2,, x n ] T Y = [y 1, y 2,, y m ] T, B = [b 1, b 2,, b m ] T and A = a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn Lecture Notes for Math 1540 First Previous Next Last 40

41 Differentials The derivative of a function f is the function denoted f and defined by f (x) = lim z x f(z) f(x) z x = lim h 0 f(x + h) f(x) h provided that the limit exists Let y = f(x) The differential of y is denoted dy and defined by dy = f (x)dx Chain rule Let y = f(x) and x = g(t), then dy dt = dy dx dx dt = f (x)g (t) Lecture Notes for Math 1540 First Previous Next Last 41

42 Integrals An antiderivative of a function f(x) is a function F (x) such that F (x) = f(x) The indefinite integral of f(x) with respect to x is defined by f(x)dx = F (x) + C, where is called the integral sign, f(x) is the integrand, x is the variable of integration, and C is the constant of integration Any two antiderivatives of a function differ only by a constant Lecture Notes for Math 1540 First Previous Next Last 42

43 Differentiation and Integration Formulas Differentiation (1) = 0 (kx) = k Integration 0dx = C kdx = kx + C (x a+1 ) = (a + 1)x a, a 1 x a dx = xa+1 a+1 + C, a 1 (ln x ) = 1 x, x 0 1 xdx = ln x + C, x 0 (e x ) = e x e x dx = e x + C (b x ) = b x (ln b), b > 0 b x dx = bx ln b + C, b > 0 (kf(x)) = kf (x) kf(x)dx = k f(x)dx (f(x) + g(x)) = f (x) + g (x) f(x) + g(x)dx = f(x)dx + g(x)dx Any two antiderivatives of a function differ only by a constant Lecture Notes for Math 1540 First Previous Next Last 43

44 The Definite Integrals Let f(x) be continuous on [a, b], then b a f(x)dx = lim n n k=1 The definite integral is the limit of an infinite sum f(a + b a n k) Fundamental theorem: Let F (x) be an antiderivative of a continuous function f(x) on [a, b], then b a f(x)dx = F (x) b a = F (b) F (a) The average of a continuous function f(x) over [a, b] is f = 1 b a b a f(x)dx Lecture Notes for Math 1540 First Previous Next Last 44

45 The Definite Integrals 1 b a (f(x) + g(x))dx = b a f(x)dx + b a g(x)dx 2 b a kf(x)dx = k b a f(x)dx 3 b a f(x)dx + c b f(x)dx = c a f(x)dx 4 b a f(x)dx = a b f(x)dx 5 b a F (x)dx = F (x) b a = F (b) F (a) Lecture Notes for Math 1540 First Previous Next Last 45

46 Areas between Curves The area of the region (vertical strip) bounded by the two curves y = f(x) and y = g(x) from x = a to x = b is b a f(x) g(x) dx The area of the region (horizontal strip) bounded by the two curves x = u(y) and x = v(y) from y = c to y = d is d c u(y) v(y) dy Lecture Notes for Math 1540 First Previous Next Last 46

47 Consumers and Producers Surplus Demand function: p = f(q) Supply function: p = g(q) Equilibrium: p 0 = f(q 0 ) = g(q 0 ) Consumers surplus: CS = q 0 0 (f(q) p 0)dq Producers surplus: P S = q 0 0 (p 0 g(q))dq Lecture Notes for Math 1540 First Previous Next Last 47

48 Methods of Integration Change of variable (let u = u(x)): f(x)dx = g(u) dx du du and b a f(x)dx = u(b) u(a) g(u) dx du du where g(u) = g(u(x)) = f(x) and dx du = (du dx ) 1 = 1 u (x) Integration by parts: uv dx = uv u vdx and b a uv dx = uv b a b a u vdx Lecture Notes for Math 1540 First Previous Next Last 48

49 Integration of Rational Functions Step 1: Long division Step 2: Factorize denominator Step 3: Partial fractions (four cases) Case 1) Case 2) Case 3) Case 4) distinct linear factors: repeated linear factors: P n 1(x) (x a) n = A 1 distinct irreducible quadratic factors: P 2n 1 (x) (x 2 +a 1 x+b 1 ) (x 2 +a n x+b n ) = A 1x+B 1 P n 1 (x) (x x 1 ) (x x n ) = A 1 x x A n x a x 2 +a 1 x+b 1 repeated irreducible quadratic factors: P 2n 1 (x) = A 1x+B 1 (x 2 +ax+b) n x 2 +ax+b + + A nx+b n (x 2 +ax+b) n A n (x a) n x x n A nx+b n x 2 +a n x+b n Lecture Notes for Math 1540 First Previous Next Last 49

50 Partial Derivatives The partial derivative of z = f(x, y) with respect to x is defined by z x = f x = f(x + h, y) f(x, y) xf(x, y) = f x (x, y) = lim h 0 h provided that the limit exists We treat y as a constant The partial derivative of z = f(x, y) with respect to y is defined by z y = f y = f(x, y + h) f(x, y) yf(x, y) = f y (x, y) = lim h 0 h provided that the limit exists We treat x as a constant Lecture Notes for Math 1540 First Previous Next Last 50

51 Competitive and Complementary Products Demand functions: q A = f(p A, p B ) and q B = g(p A, p B ) are given Competitive products (Apple and Blackberry): q A p B > 0 and q B p A > 0 Complementary products (cars and gasoline): q A p B < 0 and q B p A < 0 Remark: it is always true that q A p A < 0 and q B p B < 0 Lecture Notes for Math 1540 First Previous Next Last 51

52 Higher-Order Partial Derivatives Let z = f(x, y) f xx = (f x ) x = xx f = 2 f x 2 = x f yy = (f y ) y = yy f = 2 f y 2 = y ( ) f x ( ) f y Mixed partial derivative: f xy = f yx = 2 f x y = 2 f y x = y ( ) f x = x ( ) f y Lecture Notes for Math 1540 First Previous Next Last 52

53 Chain Rule and Implicit Partial Differentiation Let z = f(x, y) with x = x(r, s) and y = y(r, x) Then z r z s = z x x r + z y y r = z x x s + z y y s Let F (x, y, z) = 0 Then F x + F z z x = 0 By solving the above equation, we obtain z x = xf z F Lecture Notes for Math 1540 First Previous Next Last 53

54 Relative Extrema for two-variable functions The function f(x, y) has a relative maximum at the point (a, b) if f(x, y) f(a, b) for any (x, y) close to (a, b) The function f(x, y) has a relative minimum at the point (a, b) if f(x, y) f(a, b) for any (x, y) close to (a, b) Lecture Notes for Math 1540 First Previous Next Last 54

55 Relative Extrema for two-variable functions The critical points of a function f(x, y) are the solutions to the following system { f x (x, y) = 0 f y (x, y) = 0 Second-derivative test for a critical point (a, b): D(x, y) = f xx f yy f 2 xy 1) If D(a, b) > 0 and f xx (a, b) < 0, then f has a relative maximum at (a, b) 2) If D(a, b) > 0 and f xx (a, b) > 0, then f has a relative minimum at (a, b) 3) If D(a, b) < 0, then f has a saddle point at (a, b) 4) If D(a, b) = 0, then NO conclusion Lecture Notes for Math 1540 First Previous Next Last 55

56 Relative Maximum Case 1: D = f xx f yy f 2 xy > 0 and f xx < 0 Lecture Notes for Math 1540 First Previous Next Last 56

57 Relative Minimum Case 2: D = f xx f yy f 2 xy > 0 and f xx > 0 Lecture Notes for Math 1540 First Previous Next Last 57

58 Saddle Case 3: D = f xx f yy f 2 xy < 0 Lecture Notes for Math 1540 First Previous Next Last 58

59 Lagrange Multipliers Find the critical points of w = f(x 1,, x n ) subject to g(x 1,, x n ) = 0 Introduce a lagrange multiplier λ: F (x 1,, x n, λ) = f(x 1,, x n ) λg(x 1,, x n ) Find the critical points of F (x 1,, x n, λ), namely, solve the system F x 1 (x 1,, x n, λ) = f x 1 (x 1,, x n ) λ g x 1 (x 1,, x n ) = 0 F x n (x 1,, x n, λ) = f x n (x 1,, x n ) λ g x n (x 1,, x n ) = 0 F λ (x 1,, x n, λ) = g(x 1,, x n ) = 0 Lecture Notes for Math 1540 First Previous Next Last 59