Question

consider the function f(x,y)=x^2-2xy+3y^2-8y

(a)find the critical points of f and classify each critical point as local max min or saddle point

(b) does f have a global max ?if so what is it ? does f have a global min ? if so what is it ?

Answer 1

f(3) = x² - 2xy +34² - sy a point (9,6) is a critical point of f(x,y) if If(a,b) = 0 and f- of - of f. ot = 0 tx ax 2x-2y → x=y ax af 27 - 2x + by - 8 = 0 - 2x + 6y = 8 x=2 and y=2. Therefore, only critical point is (22) Now, consider, D= fxx (a, b) fry (a,b) - [ foxy (a,b)]² (1) DO and fxx > 0 relative in inamum at (a,b) and for co relative maximum at (a,b) can fxx əf . fyy = = -2 axe and ( DLO saldle at (a, b) (iv) D=0 Nothing be paid. åf f 2 ; fxy ay anay D = 2x6-(-2)² → 870 DO fox 70. letes, t(x,y) € IR? Therefore, at (2, 2) also, the above condition is true. Hence, (2,2) is relative minimum point. in, (2,2) is local minima Since only b) critical point local minima, Therefore No maxima exist. Hence, f does not have global maximum. is

global f do have minima because D> 0 and fxx> o is true t (ay) Hence, at (2,2) f attains global minimum Therefore, min 2)? - 2(2)(2) + 3(2)2 – 8(2) & fimin E-8 Hence, the minimum value is - 8.