Question

Apply a second derivative to identify a critical points as a local maximum, local minimum or saddle point for a function. Find the critical point of the function: f(x, y) = 7 + 6x - 2? + 3y + 4y? This critical point is a: Select an answer

Answer 1

Second derivative test for two variables: Suppose that f (a,b) = 0,f (a,b)= 0 and let 1 = eta 22 02 f(a,b), m= -f(a,b), n=f(a,b) ох дхду Then (i). f(a,b) is a maximum value if D= In -m>0 and 1< 0 (ii). f(a,b) is a minimum value if D= ln – m² >0 and 1> 0 (iii). f(a,b) is a daddle point. if D= In - m <0 (iv). D= ln-m²=0 then f(x,y) fails to have maximum or minimum value and it needs further investigation.

Solution: Given function is f(x,y) = 7+6x – x2 + 3y + 4y2 fx (x,y) ax a @ (7 +6X – x2 +3y +4y2) (7+6x – x2 + 3y + 4y?) f (x,y) су ff_ (x,y) =6–2x If y(x, y)=3+8y (1=f(x,y)=-2 and {m=fxy(x,y)= 0 ху n=f y(x,y)=8 Discriminent: D= Im – m² = (-2)(8) – 02 =-16 ff_(x,y)=6–2x = 0 Let for critical/saddle points. If y(x, y) = 3+8y=0 X = 3 y=-3/8 :: critical/saddle points : (x,y)= 3, 3 8 At (x,y)= 3, D=Im – m² =(-2)(8)-02 =-16<0 ..(x, y) = 3, is a saddle point.