Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3 has a relative minimum at(-2, 2)(-3, 3)(2, -2)(3, -3)(2, 2)

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Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3 has a relative minimum at(-2, 2)(-3, 3)(2, -2)(3, -3)(2, 2)

BlackTom AI · Accepted Answer

We need to analyze the function f(x,y) = 13x^3 + 12y^2 + xy − 6x + 3 and locate potential relative minima by examining critical points.

First, compute the partial derivatives:
f_x = ∂f/∂x = 39x^2 + y − 6,
f_y = ∂f/∂y = 24y + x.
Critical points occur where both partial derivatives are zero:
1) 39x^2 + y − 6 = 0
2) 24y + x = 0  ⇒  x = −24y.
Substitute x = −24y into equation (1): 39(−24y)^2 + y − 6 = 0.
Compute (−24y)^2 = 576y^2, so the equation becomes 39·576y^2 + y − 6 = 0, i.e. 22464y^2 + y − 6 = 0.
The discriminant is Δ = 1 − 4·22464·(−6) = 1 + 539,136 = 539,137, which is positive, so there are two real roots for y (and corresponding x = −24y). These critical points are therefore not simple integer coordinates; they are irrational and not among the listed candidate points.

Now, test each given option to see if it is a critical point by plugging into the gradient equations:
- For (−2, 2): f_y = 24·2 + (−2) = 48 − 2 = 46 ≠ 0, so not a critical point.
- For (−3, 3): f_y = 24·3 + (−3) = 72 − 3 = 69 ≠ 0, so not a critical point.
- For (2, −2): f_y = 24(−2) + 2 = −48 + 2 = −46 ≠ 0, so not a critical point.
- For (3, −3): f_y = 24(−3) + 3 = −72 + 3 = −69 ≠ 0, so not a critical point.
- For (2, 2): f_y = 24·2 + 2 = 48 + 2 = 50 ≠ 0, so not a critical point.

Since none of the listed points satisfy the system f_x = 0 and f_y = 0, none of them are critical points. Given that the true critical points are obtained from solving the equations above and are not among the provided options, the provided list of candidate points does not include the actual critical points.

Question at position 2 The function f(x,y)=13x3+12y2+xy−6x+3 has a relative minimum at(-2, 2)(-3, 3)(2, -2)(3, -3)(2, 2)

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