Question

Question 6. (20 pts) Find the critical points of f(x, y) = x4 + 2y2 – 4xy. Then use the Second Derivative Test to determine whether each critical point is a local min, max, or saddle point.

Answer 1

derivative test where, To find oritical points of $13) - 2497-984 akso, to determine each critical point t whether is a local min, max or saddle point using and Consider, D- fex (96) typ (a,b) - (for (as)) if, (i) do and Sex (2,670, then t has a Local minimum at (a,b). (1) Do and tre (a,b) Lo, then & has a local maximum at (a,b). (0) Dso, then f has a saddle point at (a, b). Here, f(x,y) = x1 + 277-4x7 y (7) - 123-A& shay-Ax. The critical points satisfy the equations su(x,y)-ty()) fx (2,7) = 0 and fy=0 - 44-1720 = = . 寸423-9420 put, yax, we get- Ax(x-1)=0 x=0,1,- 1 7 7 = 0, 11. [-] Critical points are (0,0); (1,1); (-1,-1) Now fox (2,4)= 120; 2x7(x,y) = 4 fry (2,4) -9. Now, at point (0,0) D-to (0,0) top (60) - Hos (0,0)) = 0.9-ig = -16 Lo. .: DCO = (0,0) is a saddle point

• at point (1,1) - 12x4 D- fox () fry (1) - Choy () (-A)" > 98-16 = 32> 0 :: Do (1,1)= 1270 So, at (1,1) t has a local minimum and for at point (-1,-1) De tuli) tyg (-4,-) - 4y (-4-1))" 12x4 - (-4) 3270 and dox (-1,-1) = 1270. So, also at (-1,-1) 7 has > local minimum