Question

4. Given the function f(x,y) = 4+x2 + y3 – 3xy. a. Find all critical points of the function. b. Use the second partials test to find any relative extrema or saddle points.

Answer 1

Second derivative test is entreme values 2 Let (a,b) be a critical point of fi B C R R, that is Df Cn 1/(a,b) = [0,0] Assume that of has Continious second dearrative in an open disr in B with Centre in (a,b) and de note B = fox (a, b) fyny (a,h) -(fay (a,h)] Off B >o and from (a,b) >0 then fla,b) is local minimim If B70 and fan (a,b) co then t(a,b) a local manimum & It Beo, then flaub) is a saddle point 6 sf B=0 the test is in conclenne. . Here B is called the discriminant of f at (a,b)

TX - 2 da 3 3 Given that of (x,y) = 4+*°+y»--371- Differentiate partially w.n-to it', y as constame. fx = load CH + x°+93_3xy) p.: d (40)=nzly = 0+38"70-34%) fx = 3x-34 0 (2120) Differentiate o partially w.97.to y',.?! as constant fy = C++ +y*_327) = 0+0 + 3y2–33201) fy = 34%3x . For coritical points, we set for=0 and fy zo 350"- 3y =0 and 34"-3= 0 to sch_yzo and go-x=0 P Å xey @ smaring on man n both sides Bc4eyrº solve ② and ③ we get 9c4-2. → 37_X =0 a 30 (x-1)=0 => x=0 (ov 1 32150 200 (ov) =1 x=0 Cortaal from ② and ② If x=0 then y=0, paint (a, gj= (0,0) If x=1 then yol point Cary) = (1,1) The critical points are (oo) and (31) a I

we have fx = 3x - 3y I we have fy = 34 3x fxx = (3x+3y) | = 3(2x) - 0 fyy sent (30°-34 ) fax = 6x l fyy = 3 (29-0=of Now fxy = y (fx) = 2 (3x² 3y) = 0-361) = -3 At the point Cool: Wow B = fxx /(a,b) - fgy | 12,5) (fey/ca,b)]? B = ( 62 )/0,0) (@y)/Co.) – [(-3)/(0) = 660) - 660) - 9 B = 0-9 =-9co By the Second derivative test, - The given function fray) has a Sadelle point at Coso). At the point (1,1): B = (64)| (11) -(6)| (11) - [(-3) Cous] = 661) » 661) -9 B = 6 x 6-9 = 36-9 =2 7 0 Now fxxl carb) = 6x1600) = 661) = 6 0 By the Second derivative test, The given function fcrry) has relative cor Local minimum at C1). Ναι