Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3f\left(x,y\right)=\frac{1}{3}x^3+\frac{1}{2}y^2+xy-6x+3 has a relative minimum at(-2, 2)(2, -2)(-3, 3)(2, 2)(3, -3)

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Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3f\left(x,y\right)=\frac{1}{3}x^3+\frac{1}{2}y^2+xy-6x+3 has a relative minimum at(-2, 2)(2, -2)(-3, 3)(2, 2)(3, -3)

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Start by clearly identifying what the problem asks: locate relative minima of f(x,y) = (1/3)x^3 + (1/2)y^2 + xy − 6x + 3 and evaluate each given option as a potential critical point.
Option 1: (-2, 2). Compute the partial derivatives: f_x = x^2 + y − 6 and f_y = y + x.
At (-2, 2): f_x = (-2)^2 + 2 − 6 = 4 + 2 − 6 = 0, and f_y = 2 + (-2) = 0, so this is a critical point. Next, examine the Hessian: f_xx = 2x, f_yy = 1, f_xy = 1. At (-2, 2), f_xx = -4, so the Hessian determinant D = f_xx f_yy − (f_xy)^2 = (-4)(1) − (1)^2 = -5 < 0. A negative determinant indicates a saddle point, not a local minimum. Therefore this option cannot be a relative minimum.
Option 2: (2, -2). Check the gradient: f_x = (2)^2 + (-2) − 6 = 4 − 2 − 6 = -4 ≠ 0, and f_y = (-2) + 2 = 0. Since the gradient is not zero, this is not a critical point at all, so it cannot be a relative minimum.
Option 3: (-3, 3). Evaluate the gradient: f_x = (-3)^2 + 3 − 6 = 9 + 3 − 6 = 6 ≠ 0, and f_y = 3 + (-3) = 0. The gradient is not zero, so this is not a critical point. Consequently, it cannot be a relative minimum.
Option 4: (2, 2). Gradient check: f_x = (2)^2 + 2 − 6 = 4 + 2 − 6 = 0, and f_y = 2 + 2 = 4 ≠ 0. Since f_y ≠ 0, this is not a critical point, hence not a relative minimum.
Option 5: (3, -3). Gradient: f_x = (3)^2 + (-3) − 6 = 9 − 3 − 6 = 0, and f_y = (-3) + 3 = 0, so this is a critical point. The Hessian components give f_xx = 2x = 6, f_yy = 1, f_xy = 1, so the determinant D = 6·1 − 1 = 5 > 0 and f_xx > 0, which confirms a local minimum at this point. Since a positive D with f_xx > 0 indicates a local minimum, this option satisfies the condition for a relative minimum.
In summary, only points where the gradient vanishes and the Hessian indicates a minimum qualify. Here, (-2, 2) is a saddle, (2,-2), (-3, 3), and (2, 2) are not critical points, and (3, -3) is a local minimum. The reasoning shows why each listed candidate fails or succeeds as a relative minimum, with the correct characterization aligning with standard second-derivative test criteria.

Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3f\left(x,y\right)=\frac{1}{3}x^3+\frac{1}{2}y^2+xy-6x+3 has a relative minimum at(-2, 2)(2, -2)(-3, 3)(2, 2)(3, -3)

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