Question

Discuss minimum value of f(x,y)=x^2 + y^2 + 6x + 12.

(a) 3

(b) 3

(c) -9

(d) 9

This question was addressed to me in my homework.

I would like to ask this question from Maxima and Minima of Two Variables in portion Maxima and Minima of Engineering Mathematics

Answer 1

Right answer is (b) 3

Explanation: Given, f(x, y) = x^2 + y^2 + 6x + 12

Now, ^∂f⁄∂x = 2x + 6 and ^∂f⁄∂y = 2

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

(x,y) = (-3,0)

Now, r= ^∂^2f⁄∂x^2 = 2>0 and t= ^∂^2f⁄∂y^2 = 2 and s= ^∂^2f⁄∂x∂y = 0

hence, rt – s^2 = 4>0 and r>0

hence. f(x,y) has minimum value at (-3,0), which is f(x,y) = 12 + 9 – 18 = 3.