Which of the following are critical points of the function $x^3-3x^2y+3y^2$?

Question

BlackTom AI · Accepted Answer

Set the gradient equal to zero: compute the partial derivatives of f(x,y) = x^3 − 3x^2y + 3y^2.

1. Relevant concepts
- A critical point occurs where the gradient (all first partial derivatives) is zero or undefined. Here f is a polynomial, so set f_x = 0 and f_y = 0.

2. Compute partial derivatives and factor
f_x = ∂/∂x (x^3 − 3x^2y + 3y^2) = 3x^2 − 6xy = 3x(x − 2y).
f_y = ∂/∂y (x^3 − 3x^2y + 3y^2) = −3x^2 + 6y = 3(2y − x^2).

Set each equal to zero:
3x(x − 2y) = 0  => x(x − 2y) = 0      (1)
3(2y − x^2) = 0 => 2y − x^2 = 0      (2)

3. Solve the system step by step
From (1), either x = 0 or x − 2y = 0.

- Case A: x = 0. Substitute into (2): 2y − 0 = 0 → y = 0. So (0,0) is a critical point.

- Case B: x − 2y = 0 → y = x/2. Substitute into (2):
2(x/2) − x^2 = x − x^2 = 0 → x(1 − x) = 0.
Thus x = 0 (gives the same (0,0) found already) or x = 1. If x = 1, y = 1/2, so (1, 1/2) is a critical point.

4. Why the correct answers are correct
Both conditions for a critical point are satisfied at (0,0) and (1,1/2), so these are the only critical points in R^2 for this polynomial function.

5. Why the other options are incorrect
- (0.5,1): f_x = 3·0.5(0.5 − 2·1) ≠ 0, so gradient ≠ 0.
- (1,1): does not satisfy 2y − x^2 = 0 because 2(1) − 1^2 = 1 ≠ 0.
- (0.5,0.5): neither x(x − 2y) nor 2y − x^2 are zero there.

Conclusion: The critical points are (0,0) and (1,0.5), matching answer choices a and c.

Which of the following are critical points of the function \(x^3-3x^2y+3y^2\)?多项选择题

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