Minima, Maxima, Saddle points
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1 Minima, Maxima, Saddle points Levent Kandiller Industrial Engineering Department Çankaya University, Turkey Minima, Maxima, Saddle points p./9
2 Scalar Functions Let us remember the properties for maxima, minima and saddle points when we have scalar functions with two variables with the help of the following examples. Minima, Maxima, Saddle points p./9
3 Scalar Functions Example. Let f(x,y) = x + y. Find the extreme points: Minima, Maxima, Saddle points p./9
4 Scalar Functions Example. Let f(x,y) = x + y. Find the extreme points: Minima, Maxima, Saddle points p./9
5 Scalar Functions Example. Let f(x,y) = x + y. Find the extreme points:.5.5 f(x,y) x = x =. f(x,y) x =, y = y =. y =. Since we have only one critical point, it is either the maximum or the minimum. We observe that f(x, y) takes only nonnegative values. Thus, we see that the origin is the minimum point. Minima, Maxima, Saddle points p./9
6 Scalar Functions Example. Find the extreme points of f(x,y) = xy x y x y + 4. Minima, Maxima, Saddle points p.3/9
7 Scalar Functions Example. Find the extreme points of f(x,y) = xy x y x y + 4. The function is differentiable and has no boundary points. Minima, Maxima, Saddle points p.3/9
8 Scalar Functions Example. Find the extreme points of f(x,y) = xy x y x y + 4. f x = f(x,y) x Thus, x = y = is the critical point. = y x, f y = f(x,y) y = x y. Minima, Maxima, Saddle points p.3/9
9 Scalar Functions Example. Find the extreme points of f(x,y) = xy x y x y + 4. f x = f(x,y) x Thus, x = y = is the critical point. f xx = f(x,y) x = y x, f y = f(x,y) y = x y. = = f(x,y) y = f yy, f xy = f(x,y) x y =. Minima, Maxima, Saddle points p.3/9
10 Scalar Functions Example. Find the extreme points of f(x,y) = xy x y x y + 4. f x = f(x,y) x Thus, x = y = is the critical point. f xx = f(x,y) x = y x, f y = f(x,y) y = x y. = = f(x,y) y = f yy, f xy = f(x,y) x y =. The discriminant (Jacobian) of f at (a, b) = (, ) is f xx f xy f xy f yy = f xx f yy f xy = 4 = 3. Since f xx <, f xx f yy f xy > f has a local maximum at (, ). Minima, Maxima, Saddle points p.3/9
11 Scalar Functions Theorem. The extreme values for f(x,y) can occur only at i. Boundary points of the domain of f. ii. Critical points (interior points where f x = f y =, or points where f x or f y fails to exist). Minima, Maxima, Saddle points p.4/9
12 Scalar Functions Theorem. If the first and second order partial derivatives of f are continuous throughout an open region containing a point (a, b) and f x (a,b) = f y (a,b) =, you may be able to classify (a,b) with the second derivative test: i. f xx <, f xx f yy fxy > at (a,b) local maximum; ii. f xx >, f xx f yy fxy > at (a,b) local minimum; iii. f xx f yy fxy < at (a,b) saddle point; iv. f xx f yy fxy = at (a,b) test is inconclusive (f is singular). Minima, Maxima, Saddle points p.4/9
13 Quadratic forms Definition. The quadratic term f(x, y) = ax + bxy + cy is positive definite (negative definite) if and only if a > (a < ) and ac b >. Minima, Maxima, Saddle points p.5/9
14 Quadratic forms Definition. The quadratic term f(x, y) = ax + bxy + cy is positive definite (negative definite) if and only if a > (a < ) and ac b >. f has a minimum (maximum) at x = y = if and only if f xx (, ) > (f xx (, ) < ) and f xx (, )f yy (, ) > f xy(, ). Minima, Maxima, Saddle points p.5/9
15 Quadratic forms Definition. The quadratic term f(x, y) = ax + bxy + cy is positive definite (negative definite) if and only if a > (a < ) and ac b >. f has a minimum (maximum) at x = y = if and only if f xx (, ) > (f xx (, ) < ) and f xx (, )f yy (, ) > f xy(, ). If f(, ) =, we term f as positive (negative) semi-definite provided the above conditions hold. Minima, Maxima, Saddle points p.5/9
16 Quadratic forms Now, we are able to introduce matrices to the quadratic forms: ax + bxy + cy = [x,y] a b x. b c y Minima, Maxima, Saddle points p.6/9
17 Quadratic forms Thus, for any symmetric A, the product f = x T Ax is a pure quadratic form: it has a stationary point at the origin and no higher terms. Minima, Maxima, Saddle points p.6/9
18 Quadratic forms Thus, for any symmetric A, the product f = x T Ax is a pure quadratic form: it has a stationary point at the origin and no higher terms. a a a n x xa T a a a n x x = [x,x,,x n ] a n a n a nn x n Minima, Maxima, Saddle points p.6/9
19 Quadratic forms Thus, for any symmetric A, the product f = x T Ax is a pure quadratic form: it has a stationary point at the origin and no higher terms. a a a n x xa T a a a n x x = [x,x,,x n ] a n a n a nn x n = a x + a x x + + a nn x n = n n i= j= a ijx i x j. Minima, Maxima, Saddle points p.6/9
20 Quadratic forms Definition. If A is such that a ij = f x i x j (hence symmetric), it is called the Hessian matrix. Minima, Maxima, Saddle points p.7/9
21 Quadratic forms Definition. If A is such that a ij = f x i x j (hence symmetric), it is called the Hessian matrix. If A is positive definite (x T Ax >, x θ) and if f has a stationary point at the origin (all first derivatives at the origin are zero), then f has a minimum. Minima, Maxima, Saddle points p.7/9
22 Quadratic forms Remark. Let f : R n R and x R n be the local minimum, f(x ) = θ and f(x ) is positive definite. Minima, Maxima, Saddle points p.8/9
23 Quadratic forms Remark. Let f : R n R and x R n be the local minimum, f(x ) = θ and f(x ) is positive definite. We are able to explore the neighborhood of x by means of x + x, where x is sufficiently small (such that the second order Taylor s approximation is pretty good) and positive. Minima, Maxima, Saddle points p.8/9
24 Quadratic forms Remark. Let f : R n R and x R n be the local minimum, f(x ) = θ and f(x ) is positive definite. We are able to explore the neighborhood of x by means of x + x, where x is sufficiently small (such that the second order Taylor s approximation is pretty good) and positive. Then, f(x + x) = f(x ) + x T f(x ) + xt f(x ) x. Minima, Maxima, Saddle points p.8/9
25 Quadratic forms Remark. Let f : R n R and x R n be the local minimum, f(x ) = θ and f(x ) is positive definite. We are able to explore the neighborhood of x by means of x + x, where x is sufficiently small (such that the second order Taylor s approximation is pretty good) and positive. Then, f(x + x) = f(x ) + x T f(x ) + xt f(x ) x. The second term is zero since x is a critical point and the third term is positive since the Hessian evaluated at x is positive definite. Thus, the left hand side is always strictly greater than the right hand side, indicating the local minimality of x. Minima, Maxima, Saddle points p.8/9
26 E Collaborative Work: Let f(x, x ) = 3 x3 + x + x x + x x + 9. Find the stationary and boundary points, then find the minimizer and the maximizer over 4 x x ,,4,6,8,,4,6,8 C A,,4 B,6, ,5 -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
27 E Collaborative Work: Let f(x, x ) = 3 x3 + x + x x + x x + 9. Find the stationary and boundary points, then find the minimizer and the maximizer over 4 x x ,,4,6,8,,4,6,8 C A, E,4 B,6, ,5 -,9 -,4 -,9 -, ,,4,6,8,,4,6,8 C A B,,4,6, ,4 -,9 -,4 -,9-3,5 Minima, Maxima, Saddle points p.9/9
28 E Collaborative Work: Let f(x, x ) = 3 x3 + x + x x + x x + 9. Find the stationary and boundary points, then find the minimizer and the maximizer over 4 x x 3 f x f(x) = = x + x + x =. (x )(x ) = x + x x = x f x ,,4,6,8,,4,6,8 C A,,4 B,6, ,5 -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
29 E 33 3 Collaborative Work: f(x) = f x f x = x + x + x x + x. = ,,4,6 C A B,8,,4,6,8,,4,6, ,4 -,9 -,4 -,9-3,5-4 (x )(x ) = x = x Therefore, x A = ¾, xb = defined by 4 x x 3. ¾ 3 are stationary points inside the region Minima, Maxima, Saddle points p.9/9
30 E 33 3 Collaborative Work: f(x) = f x f x = x + x + x x + x. = ,,4,6 C A B,8,,4,6,8,,4,6, ,4 -,9 -,4 -,9-3,5-4 (x )(x ) = x = x Therefore, x A = ¾, xb = defined by 4 x x 3. ¾ 3 are stationary points inside the region Moreover, we have the following boundaries ¾ defined by x I = ¾ x x C = ¾, xii =, xd = ¾ 3 x ¾ and xiii = 4, xe = ¾ x 4 3, xiv =, xf = ¾ x 3 4. ¾ Minima, Maxima, Saddle points p.9/9
31 E Collaborative Work: Let the Hessian matrix be f(x) = f x x f x x f x x f x x = x ,,4,6,8.,,4,6,8 C A,,4 B,6, ,4 -,9 -,4 -,9-3,5 Minima, Maxima, Saddle points p.9/9
32 E Collaborative Work: Let the Hessian matrix be f(x) = f x x f x x have f(x A ) = f x x f x x 3 = x , and f(x B ) =,4,6,8,,4,6,8 C A,,4 B,6, ,5. Then, we 5. -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
33 E Collaborative Work: Let the Hessian matrix be f(x) = f x x f x x have f(x A ) = f x x f x x 3 = x , and f(x B ) = Let us check the positive definiteness of f(x A ): v T f(x A )v = [v, v ] ¾ 3 ¾ v v,4,6,8,,4,6,8 C A,,4 B,6, ,5. Then, we 5. = 3v + 4v v + v. If v =.5 and v =., we will have v T f(x A )v <. On the other hand, if v =.5 and v =., we will have v T f(x A )v >. Thus, f(x A ) is indefinite. -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
34 E Collaborative Work: Let the Hessian matrix be f(x) = f x x f x x have f(x A ) = Let us check f(x B ): v T f(x B )v = [v, v ] f x x f x x 3 ¾ 5 = x , and f(x B ) = ¾ v v Thus, f(x B ) is positive definite and x B = f(x B ) = ,4,6,8,,4,6,8 C A,,4 B,6, ,5. Then, we 5. = 5v + 4v v + v = v + (v + v ) >. ¾ 3 is a local minimizer with -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
35 E Collaborative Work: C A B ,4 -,9 -,4 Let us check the boundary defined by x I :,,4,6,8,,4,6,8,,4,6, ,9-3,5 f(, x ) = x x + 9 df(, x ) dx = x. = x =. Since d f(,x ) = >, x dx = > is the local minimizer outside the feasible region. As the first derivative is negative for 4 x, we will check x = for minimizer and x = 4 for maximizer. Minima, Maxima, Saddle points p.9/9
36 E Collaborative Work: C A B ,4 -,9 -,4 Let us check the boundary defined by x II :,,4,6,8,,4,6,8,,4,6, ,9-3,5 f(3, x ) = x + 5x + 65 df(3, x ) dx = x + 5. = x = 5. Since d f(,x ) = >, x dx = 5 < 4 is the local minimizer outside the feasible region. As the first derivative is positive for 4 x, we will check x = 4 for minimizer and x = for maximizer. Minima, Maxima, Saddle points p.9/9
37 E Collaborative Work: C A B ,4 -,9 -,4 Let us check the boundary defined by x III :,,4,6,8,,4,6,8,,4,6, ,9-3,5 f(x,) = 3 x3 + x +9 df(x,) dx = x +x. = x =,. Since d f(x,) dx ( d f(,) dx = x +, x = is the local minimizer = > ) on the boundary, and x = is the local maximizer ( d f(,) = < ) outside the feasible region. As dx the first derivative is positive for x 3, we will check x = 3 for maximizer. Minima, Maxima, Saddle points p.9/9
38 E Collaborative Work: Let us check the boundary defined by x IV : f(x, 4) = 3 x3 + x 8x + 3 df(x, 4) dx = x + x 8 =. x = ± +3. Since d f(x, 4) dx root x = + 33 =.373 is the local minimizer ( d f(.373,) dx ,,4,6,8, = x + again, the positive > ), and the negative root is the local maximizer but it is outside the feasible region. As the first derivative is positive for x 3, we will check x = 3 for maximizer again.,4,6,8 C A,,4 B,6, ,5 -,9 -,4 -,9 -, Minima, Maxima, Saddle points p.9/9
39 E Collaborative Work: C A B ,4 -,9 -,4,,4,6,8,,4,6,8,,4,6, ,9-3,5 To sum up, we have to consider (, 3), (,) and (.373, 4) for the minimizer; (3,) and (, 4) for the maximizer: f(, 3) = , f(,) = 9, f(.373, 4) = (,) is the minimizer! f(3,) = 3.5, f(, 4) = 3 (3,) is the maximizer! Minima, Maxima, Saddle points p.9/9
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