multivariate optimisation -

Question

Given f(x,y) = x³+y³-10xy+25, S = {(x,y): 0 ≤ x ≤ 6, 0 ≤ y ≤ 6}

match the point on the left with the classification on the right:

(0,6)

(10/3, 10/3)

(Square root(20), 6)

Global minimum

Neither

Global max

Answer 1

f(x,y) = x³+y³-10xy+25

∂f/∂x = 3x²-10y

∂f/∂y = 3y²-10x

3x²-10y = 0

3x² = 10y

x = √(10y/3) <-- x has to be positive, so -√(10y/3) isn't considered

3y²-10x = 0

3y²-10√(10y/3) = 0

3y² = 10√(10y/3)

9y⁴ = 100(10y/3)

9y⁴ = 1000y/3

27y⁴ - 1000y = 0

y(27y³-1000) = 0

y(3y-10)(9y²+30y+100) = 0 <-- Use difference of cubes formula

y₁=0

3y-10 = 0 --> y₂=10/3

9y²+30y+100 = 0 has a complex solution because its discriminant is <0, so it's not considered

List critical points

(x,y₁) = √(10(0)/3) = (0,0)

(x,y₂) = √(10(10/3)/3) = (10/3,10/3) <-- again, x must be positive, so x=-10/3 is not considered

Determine 2nd partial derivatives for Hessian matrix

∂²f/∂x² = 6x

∂²f/∂y² = 6y

∂²f/∂x∂y = -10

At (0,0):

H = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = [6(0)][6(0)] - (-10)² = 0 - 100 = -100<0, so (0,0) is a saddle point, which means (0,6) is also a saddle point, which means (0,6) is neither

At (10/3,10/3):

H = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = [6(10/3)][6(10/3)] - (-10)² = (20)(20) - 100 = 400 - 100 = 300>0, so (10/3,10/3) is either a local maximum or minimum, so it's not global, which means it's neither

At (√20,6):

H = (∂²f/∂x²)(∂²f/∂y²) - (∂²f/∂x∂y)² = (6√20)[6(6)] - (-10)² = 216√20 - 100 > 0, so (√20,6) is a global minimum because ∂²f/∂x² > 0.

Hope this helped!