Answer:
Step-by-step explanation:
[tex](a) f(x, y) = x^ 2-3xy + y ^2 + 2x + 5[/tex]
Let us find out partial derivatives
[tex]f_x =2x-3y+2\\f_y = -3x+2y[/tex]
Equate these to 0 and solve
we get
[tex]6x-9y+6 =0\\-6x+4y=0\\-5y+6 =0\\y=1.2 \\x=0.8[/tex]
Critical point (0.8,1.2)
b) [tex]f(x, y) = x^ 2 - 2xy + y^ 2\\f_x=2x-2y=0\\f_y = -2x+2y =0\\x=y[/tex]
So (x=y) is the solution
c) [tex]f(x, y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F\\f_x=2Ax+By+D\\f_y=Bx+2cY+E[/tex]
Equate to 0 and solve
2ABx + B^2y +DB =0
2ABx+4ACy +2AE =0
[tex]y =\frac{BD-2AE}{B^2-4AC}[/tex]
Similarly for x
4ACX + 2BCy +2CD =0
B^2 x +2BCy+BE=0
[tex]x=\frac{BE-2CD}{B^2-4AC}[/tex]
