Question

(c) Determine the critical points of the functions below and find out whether each point corresponds to a relative minimum, maximum, saddle point or no conclusion can be made. f(x,y) = 3xy - x - y (7 marks)

Answer 1

Lat fcz y) = 3xy-x3-43 be calculate Critical point nond of(2)= x (x,y) = the given function. of fray)= 3xy -x3-y} Se cay) is partial derivative of of (2,y) with respect to a. 34-322 7 and afcab) = fy (ay) = 34-3y2 zy fy cay is partial derivative fix,y) with respect to y he Solve the for critical pointi function fxcx,y) = fy (854)=0 tx (x,y)=0 ² 34-32²=0 y=x² and tycky)=0 ) 3x-3y²= => x=y2 = y² on and get Put x=y2 y = (y22 > y=ya => 4(y 3-1)=0 > y=0 othereuuse OY y=1, values of y is imaginary, 20 then azo Set 32= 11. when and when yot then

х9 mo again х-ч? эх= 1, ч n=nd then y=1 this not real naheeda so х+ - . therefore critical points co,o) and (+,2). xx (ҳо) =- 6x fx2(x,y) is partial derivative of fu cay) with respect to x. and $ay (9) = - - 64 tyg (ky) is partial derivative with respect to y. - р (Sa). 4X-2) - х = 3 day is partial derivative of fy with respect 2 хх ач - к 1 са and tynn (ky) = fyz=3 tya is partial derivative of ta noortoy

at the point (0,0) D= textyy-tay fxx = fxx (82) , tyy= fgy (2), the values of D D = -6%. -67 – 3) = 36*y-9, coo) = -9 <0, saddle point. ol so (oo) be and values of Dat of D at the point (t, x) تدري اه 36.d.t-9 -2730, (1,1) maximum .08 the potent so minimum be (t, 2) point. -6X. now and Hxx = fyy=-by. F-6 KO and fxx lett = -6 KO, fyy lett) thence (4,1) is the Point that aftains manim um vahee of fixya = 3xy - x3y3. [14-D 10.1 = 3-1-1=1 1. coo be the saddle point. (4,7) be the maximum point.