Question

Discuss maximum or minimum value of f(x,y) = y^2 + 4xy + 3x^2 + x^3.

(a) minimum at (0,0)

(b) maximum at (0,0)

(c) minimum at (2/3, -4/3)

(d) maximum at (2/3, -4/3)

I got this question in an online interview.

Question is from Maxima and Minima of Two Variables in portion Maxima and Minima of Engineering Mathematics

Answer 1

The correct choice is (c) minimum at (2/3, -4/3)

The best I can explain: Given,f(x,y) = y^2 + 4xy + 3x^2 + x^3

Now,^∂f⁄∂x = 4y + 6x + 3x^2 and ^∂f⁄∂y = 2y + 4x

Putting,^∂f⁄∂x and ^∂f⁄∂y = 0,and solving two equations,we get,

(x,y) = (0,0) or (2/3, -4/3)

Now,at (0,0) r= ^∂^2f⁄∂x^2=6+6x=6>0 and t= ^∂^2f⁄∂y^2 =2>0 and s= ^∂^2f⁄∂x∂y=4

hence, rt – s^2 = 12 – 16<0,hence it has no extremum at this point.

Now at (^2⁄3,-^4⁄3) r=^∂^2f⁄∂x^2= 6 + 6x = 10>0 and t= ^∂^2f⁄∂y^2 =2>0 and s= ^∂^2f⁄∂x∂y=4

hence, rt – s^2 = 20 – 16 > 0 and r>0, hence it has minimum at this point.(^2⁄3, –^4⁄3).