Find the maxima, minima, and saddle points of the function f(x,y) = 4xy - x4 - y4.

To find the maxima, minima, and saddle points of the function f(x,y) = 4xy - x^4 - y^4, we follow these steps: Step 1: Find the first partial derivatives and set them to zero to find critical points. The given function is f(x,y) = 4xy - x^4 - y^4. The partial derivative with respect to x is: f_x = ()/( x)(4xy - x^4 - y^4) = 4y - 4x^3 The partial derivative with respect to y is: f_y = ()/( y)(4xy - x^4 - y^4) = 4x - 4y^3 Set f_x = 0 and f_y = 0: 1) 4y - 4x^3 = 0 y = x^3 2) 4x - 4y^3 = 0 x = y^3 Substitute equation (1) into equation (2): x = (x^3)^3 x = x^9 x^9 - x = 0 x(x^8 - 1) = 0 This gives three possible values for x: • x = 0 • x^8 - 1 = 0 x^8 = 1 x = ± 1 Now, find the corresponding y values using y = x^3: • If x = 0, then y = 0^3 = 0. Critical point: (0,0). • If x = 1, then y = 1^3 = 1. Critical point: (1,1). • If x = -1, then y = (-1)^3 = -1. Critical point: (-1,-1). The critical points are (0,0), (1,1), and (-1,-1). Step 2: Find the second partial derivatives. f_xx = ()/( x)(4y - 4x^3) = -12x^2 f_yy = ()/( y)(4x - 4y^3) = -12y^2 f_xy = ()/( y)(4y - 4x^3) = 4 Step 3: Calculate the discriminant D(x,y) = f_xxf_yy - (f_xy)^2. D(x,y) = (-12x^2)(-12y^2) - (4)^2 D(x,y) = 144x^2y^2 - 16 Step 4: Apply the Second Derivative Test to each critical point. At (0,0): D(0,0) = 144(0)^2(0)^2 - 16 = -16 Since D(0,0) < 0, the point (0,0) is a saddle point. At (1,1): D(1,1) = 144(1)^2(1)^2 - 16 = 144 - 16 = 128 Since D(1,1) > 0, we check f_xx(1,1): f_xx(1,1) = -12(1)^2 = -12 Since f_xx(1,1) < 0, the point (1,1) is a local maximum. The value of the function at this point is f(1,1) = 4(1)(1) - (1)^4 - (1)^4 = 4 - 1 - 1 = 2. At (-1,-1): D(-1,-1) = 144(-1)^2(-1)^2 - 16 = 144 - 16 = 128 Since D(-1,-1) > 0, we check f_xx(-1,-1): f_xx(-1,-1) = -12(-1)^2 = -12 Since f_xx(-1,-1) < 0, the point (-1,-1) is a local maximum. The value of the function at this point is f(-1,-1) = 4(-1)(-1) - (-1)^4 - (-1)^4 = 4 - 1 - 1 = 2. Summary of results: • The point (0,0) is a saddle point. • The point (1,1) is a local maximum with value f(1,1) = 2. • The point (-1,-1) is a local maximum with value f(-1,-1) = 2. The final answer is Saddle point at (0,0), Local maxima at (1,1) and (-1,-1) with value 2. Send me the next one 📸