This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Step 1: Find the first-order partial derivatives f_x and f_y. The given function is: f(x,y) = x^3 + 3xy^2 - 15x + y^3 - 15y Differentiate f(x,y) with respect to x (treating y as a constant): f_x = ()/( x)(x^3 + 3xy^2 - 15x + y^3 - 15y) f_x = 3x^2 + 3y^2 - 15 Differentiate f(x,y) with respect to y (treating x as a constant): f_y = ()/( y)(x^3 + 3xy^2 - 15x + y^3 - 15y) f_y = 6xy + 3y^2 - 15 Step 2: Find the critical points by setting f_x = 0 and f_y = 0. Set f_x = 0: 3x^2 + 3y^2 - 15 = 0 Divide by 3: x^2 + y^2 - 5 = 0 (1) Set f_y = 0: 6xy + 3y^2 - 15 = 0 Divide by 3: 2xy + y^2 - 5 = 0 (2) From equation (1), we have x^2 + y^2 = 5. From equation (2), we have 2xy + y^2 = 5. Equating the two expressions for 5: x^2 + y^2 = 2xy + y^2 x^2 = 2xy x^2 - 2xy = 0 Factor out x: x(x - 2y) = 0 This gives two cases: Case 1: x = 0 Substitute x = 0 into equation (1): 0^2 + y^2 - 5 = 0 y^2 = 5 y = ±sqrt(5) This gives two critical points: (0, sqrt(5)) and (0, -sqrt(5)). Case 2: x - 2y = 0 x = 2y Substitute x = 2y into equation (1): (2y)^2 + y^2 - 5 = 0 4y^2 + y^2 - 5 = 0 5y^2 - 5 = 0 5y^2 = 5 y^2 = 1 y = ± 1 If y = 1, then x = 2(1) = 2. Critical point: (2, 1). If y = -1, then x = 2(-1) = -2. Critical point: (-2, -1). The critical points are (0, sqrt(5)), (0, -sqrt(5)), (2, 1), and (-2, -1). Step 3: Calculate the second-order partial derivatives. f_xx = ()/( x)(3x^2 + 3y^2 - 15) = 6x f_yy = ()/( y)(6xy + 3y^2 - 15) = 6x + 6y f_xy = ()/( y)(3x^2 + 3y^2 - 15) = 6y Step 4: Calculate the discriminant D(x,y). D(x,y) = f_xxf_yy - (f_xy)^2 D(x,y) = (6x)(6x + 6y) - (6y)^2 D(x,y) = 36x^2 + 36xy - 36y^2 D(x,y) = 36(x^2 + xy - y^2) Step 5: Evaluate D(x,y) and f_xx(x,y) at each critical point. For (0, sqrt(5)): D(0, sqrt(5)) = 36(0^2 + 0(sqrt(5)) - (sqrt(5))^2) = 36(0 + 0 - 5) = -180 Since D(0, sqrt(5)) < 0, (0, sqrt(5)) is a saddle point. The function value is f(0, sqrt(5)) = 0^3 + 3(0)(sqrt(5))^2 - 15(0) + (sqrt(5))^3 - 15(sqrt(5)) = 5sqrt(5) - 15sqrt(5) = -10sqrt(5). For (0, -sqrt(5)): D(0, -sqrt(5)) = 36(0^2 + 0(-sqrt(5)) - (-sqrt(5))^2) = 36(0 + 0 - 5) = -180 Since D(0, -sqrt(5)) < 0, (0, -sqrt(5)) is a saddle point. The function value is f(0, -sqrt(5)) = 0^3 + 3(0)(-sqrt(5))^2 - 15(0) + (-sqrt(5))^3 - 15(-sqrt(5)) = -5sqrt(5) + 15sqrt(5) = 10sqrt(5). For (2, 1): D(2, 1) = 36(2^2 + (2)(1) - 1^2) = 36(4 + 2 - 1) = 36(5) = 180 Since D(2, 1) > 0, we check f_xx(2, 1): f_xx(2, 1) = 6(2) = 12 Since f_xx(2, 1) > 0 and D(2, 1) > 0, (2, 1) is a local minimum. The function value is f(2, 1) = 2^3 + 3(2)(1)^2 - 15(2) + 1^3 - 15(1) = 8 + 6 - 30 + 1 - 15 = 15 - 45 = -30. For (-2, -1): D(-2, -1) = 36((-2)^2 +