multivariate optimisation - Given f(x,y) = x^3 - 3x + xy + y^2,

Question

the saddle point is ( , ) and the local minimum is ( , ). Round your answer to 4 decimal places. 

Answer 1

Determine the critical points (a,b) such that the partial derivatives ∂f/∂x (a,b) = ∂f/∂y (a,b) = 0:

∂f/∂x = 3x² - 3 + y

∂f/∂y = y + 2y = 3y

3x² - 3 + y = 0

3y = 0 --> y = 0

3x² - 3 = 0

x² - 1 = 0

(x-1)(x+1) = 0

x = -1,1

Therefore, the critical points of the multivariable function are (-1,0) and (1,0).

To get the saddle point and local minimum, we find the determinant of the Hessian matrix at those critical points:

H = (∂²f/∂x∂x)(∂²f/∂y∂y) - (∂²f/∂x∂y)²

H = (6x)(3) - 1²

H = 18x + 2

At (1,0):

H = 18(1) + 2 = 18 + 2 = 20>0, so (1,0) is either a local maximum or minimum. Hence, we have to check the value of ∂²f/∂x∂x at (1,0). Because ∂²f/∂x∂x = 18(1) = 18>0, then (1,0) is a local minimum.

At (-1,0):

H = 18(-1) + 2 = -18 + 2 = -16<0, so (-1,0) is a saddle point.

Hope this helped!