Question

Find and classify all critical points of f(x,y)=cosx+cosy. Find and classify all the critical points of f(x,y)= x^4y^4-4xy.

Answer 1

The function f(x, y) = x⁴y⁴ - 4xy

f_x  = 4x³y⁴ - 4y

f_y  = 4x⁴y³ - 4x

 

Solve f_x = 0 and f_y  = 0

 4x³y⁴ - 4y = 0 ---> (1)  4x⁴y³ - 4x = 0 ---> (2)

From first equation 4y(x³y³ - 1) = 0

Apply zero product property.

4y = 0 and x³y³ - 1 = 0

y = 0 and x³y³ = 1

From second equation, 4x(x³y³ - 1) = 0

Apply zero product property.

4x = 0 and x³y³ - 1 = 0

x = 0 and x³y³ = 1

The only intersecting point is (0, 0).

Critical point is (0, 0).

f_xx = 12x²y⁴

f_yy = 12x⁴y²

f_xy = 16x³y³ - 4

f_xx at (0, 0) = 0, f_yy at (0, 0)= 0 and f_xy at (0, 0) = - 4

The discriminant f_xx f_yy - f²_xy at a critical point P(x₀ , y₀) = (0, 0).

= (0) (0) - (- 4)²

= - 16

In this case f_xx f_yy - f²_xy is negative.

The discriminant is less than zero, then f has neither a local minimum nor a local maximum at (0, 0).

Answer 2

The function f(x, y) = cosx + cosy

f_x = - sinx

f_y = - siny

Solve f_x = 0 and f_y  = 0

- sinx = 0 and - siny = 0

sinx = 0 and  siny = 0

The general solution of sinθ = 0  then θ = nπ where n is an integer.

θ = x

 x = nπ

θ = y

y = nπ

f_xx = - cosx and f_yy = - cosy

f_xy = 0

f_xx at (nπ, nπ) = - cos(nπ), f_yy at (nπ, nπ) = - cos(nπ) and f_xy at (nπ, nπ) = 0

f_xx at (nπ, nπ) = - cos(nπ)

For n = 0, - cos(0) = - 1 < 0

For n = 1, - cos(π) =  1 > 0.

 

The discriminant f_xx f_yy - f²_xy at a critical point P(x₀ , y₀) = (nπ, nπ).

= (- cos(nπ))(- cos(nπ)) - 0

= cos²(nπ)

For n = 0

cos²(0) = 1

For n = 1

cos²(π) = 1

For n = 2

cos²(2π) = 1

In this case f_xx f_yy - f²_xy is positive.

The discriminant is greater than zero and f_xx  > 0 then f has local minimum.

The discriminant is greater than zero and f_xx  < 0 then f has local maximum.

The critical point is (nπ, nπ).