critical points: 1,1, 1, 1, 1,1, 1, 1

Transcription

1 Extrema of Functions of Several Variables -(.7). Local Extremes and Critical Points Definition: The point a,b is a critical point of a function f x,y if a,b is in the domain of f and either f a,b or f a,b DNE one or both of f x a,b and f y a,b don t exist) A necessary condition of a local extremum: If f x,y has a local extremum at a,b, then a,b must be a critical point of f. Warning!! A critical point of f may not be a local extremum. Saddle points: The point a,b, f a,b is a saddle point of z f x,y if a,b is a critical point of f andifeveryopendiskcenteredat a,b in the domain of f contains points x,y for which f x,y f a,b and points x,y for which f x,y f a,b. Example Let f x,y xe x / y 3 /3 y. Find all critical points of f and determine graphically if they are local extrema. f x e x / y 3 /3 y x x e x / y 3 /3 y f y x y e x / y 3 /3 y x x or y x and x ory critical points:,,,,,,,

2 Second Derivatives Test Suppose that f x,y has continuous second-order partial derivatives in some open disk containing the point a,b, and f x a,b f y a,b.definethe discriminant D for the point a,b by D a,b f xx a,b f yy a,b f xy ab. a. If D a,b, and f xx a,b, then f has a local minimum at a,b. b. If D a,b, and f xx a,b, then f has a local maximum at a,b. c. If D a,b, then f has a saddle point at a,b. d. If D a,b, then no conclusion can be drawn from this test. Example Let f x,y x 3 y y 3x y. Locate and classify all critical points for f x,y.

3 f x 3x xy f y y 8y 3 3x, f x 3x x y, x orx y When x, f y y 8y 3 y y, y When x y, f y y 8y 3 3 y y 8y 3 y y 3y y y y y x x critical points:,,,,, 3

4 f xx x y, f yy y, f xy x D x,y x y y x x y y 3x D,, D,, D,, f xx, Hence, from the Second Derivatives Test, we know, is a saddle point,, is a local maximum point, and there is no conclusion at,. Now let us check out the case at,. When y, f x, x 3, x is an inflection point. When x, f,y y y,a maximum point. Hence,, is neither a local maximum nor a local minimum point. y, - - x 3. Absolute maximum and minimum Extreme value Theorem: Let f x,y be continuous on a closed and bounded region R. Then f must has both an absolute maximum and minimum on R. Further more, the absolute extrema must occur at either a critical

5 point in R or on the boundary of R. Example Find the absolute extreme values of f x,y 5 x x 3y y on the region R bounded by lines y, y x and y x. a. Find all local extremes. Find all critical points: f x,y x, 3 y, x, x 3 y, y 3 critical point:, 3 f, b. Find the absolute maximum and minimum on the boundary: y, y x and y x. y This boundary consists of 3 pieces. C : y x, x, g x f x,x 5 x x 3x x 5 7x 3x y g x x 7, x 7, g g 5, g 5 7 C : y, x, g x f x, 5 x x 3 7 x x x x 5

6 y x g x x x, x g 7 9, g , g C 3 : y x, x, g x f x, x 5 x x 3 x x 5 x 3x x y f x x, x is not in,. f 5 9, f 5 a,b f a,b Conclusion critical point, absolute maximum boundary y x, boundary y, 9, 9 absolute minimum boundary y x, 5, 9 Conclusion: The absolute maximum value is 9.5 at, 3 and absolute minimum value is 9at,.

7 Steepest Ascent Method and Steepest Descent Method: Recall that f x,y changes the most at a,b along the direction u f a,b. At a,b, the line in the x t a f x a,b t direction f a,b is. We would like to find t such that y t b f y a,b t g t f x t,y t changes the most. We first find t such that g t and then find the maximum of g t c, g and g. Steps: () Compute f a,b. () Set parametric equations for the line: x t a f x a,b t y t b f y a,b t (3) Set g x t,y t and find all critical numbers t c such that g t c. Note that g t f x x t,y t x t f y x t,y t y t. () Choose t from t c,, such that g t max g t c, g, g. (5) Compute the new step: a new,b new x t, y t Example: f x,y xy x y. Carry out one step by the Method of Steepest Ascent at,3. () f x,y y x 3,x y 3, f,3 3 3, 3 3, x t t () Line: y t 3 t TI-89: at HOME, type in t,presssto, type in x t, enter type in 3 t,presssto, type in y t, enter 7

8 (3) g t f x t,y t t 3 t t 3 t TI-89: at HOME type in x t y t x t y t, press STO, type in g t, enter (i) Critical number: g t y x 3 x y 3 y x 3 5x 5y 3 y x 3 5x 5y 3 TI-89: at HOME F/Solve, type in y t x t 3 5x t 5 y t 3,t, green/enter t. 5 type in.5, press STO, type in tb, enter (ii) Check: g 9., g , g TI-89: at HOME type in g, enter type in g, enter type in g tb,enter t tb. 5 () new step: x y f.599, TI-89: at HOME type in x tb type in y tb 8