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)(-2, 2)(-3, 3)(3, -3)Single choice
A
(2, -2)
B
(2, 2)
C
(-2, 2)
D
(-3, 3)
E
(3, -3)

Question

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)(-2, 2)(-3, 3)(3, -3)

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First, I will restate the problem clearly and list the answer choices so we can analyze each option in turn. Question: The function f(x,y) = (1/3) x^3 + (1/2) y^2 + x y − 6 x + 3 has a relative minimum at which point? Choices: (2, −2), (2, 2), (−2, 2), (−3, 3), (3, −3). Next, I will consider where the gradient is zero to locate critical points. The partial derivatives are: f_x = ∂/∂x [(1/3)x^3 + (1/2)y^2 + x y − 6 x + 3] = x^2 + y − 6, f_y = ∂/∂y [(1/3)x^3 + (1/2)y^2 + x y − 6 x + 3] = y + x. Setting them equal to zero gives the system: x^2 + y − 6 = 0, y + x = 0. From the second equation, y = −x. Substituting into the first, x^2 − x − 6 = 0, which factors as (x − 3)(x + 2) = 0, so x = 3 or x = −2. Therefore the candidate critical points are: (x, y) = (3, −3) and (−2, 2). Now I analyze the nature of these critical points via the Hessian. The second partial derivatives are: f_xx = ∂^2 f/∂x^2 = 2x, f_yy = ∂^2 f/∂y^2 = 1, f_xy = f_yx = ∂^2 f/∂x∂y = 1. The Hessian matrix is H = [[2x, 1], [1, 1]]. The determinant D = f_xx·f_yy − (f_xy)^2 = (2x)(1) − 1 = 2x − 1. For the point (3, −3): f_xx = 2·3 = 6, f_yy = 1, D = 2·3 − 1 = 5 > 0 and f_xx > 0, which indicates a local minimum at (3, −3). For the point (−2, 2): f_xx = 2·(−2) = −4, f_yy = 1, D = 2·(−2) − 1 = −5 < 0, which indicates a saddle point at (−2, 2). With this information, we can evaluate each option: - (2, −2): This point is not even a critical point since it does not satisfy y = −x (here y would need to be −2 while x is 2, but the gradient equations do not vanish there). Consequently, it cannot be a relative min. It also does not satisfy the system x^2 + y − 6 = 0 and y + x = 0, so it fails at the first step of critical-point testing. - (2, 2): Substituting into the gradient: f_y = y + x = 4 ≠ 0, so not a critical point; thus cannot be a relative min. - (−2, 2): This is one of the critical points found, but the Hessian analysis shows D < 0, i.e., a saddle point, not a local minimum. - (−3, 3): Substituting into the gradient: f_y = y + x = 0, but f_x = x^2 + y − 6 = 9 + 3 − 6 = 6 ≠ 0, so not a critical point; hence cannot be a relative min. - (3, −3): This point does satisfy both gradient equations and, as shown, yields D > 0 with f_xx > 0, confirming a local minimum. In summary, the only point that passes the full test—being a critical point and yielding a local minimum via the Hessian—is (3, −3). Although some of the other options might appear plausible at a glance if one only loosely checks the gradient, the rigorous test shows they are either not critical points or correspond to saddles rather than minima.

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)(-2, 2)(-3, 3)(3, -3)Single choice

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