Questions: At least one of the answers above is NOT correct. Consider the function f(x, y)=x^2 y+y^3-48 y f has some other critical point v at (0,-4). f has ? at (-4 sqrt(3), 0) f has ? at (4 sqrt(3), 0) f has ? at (0,0) f has ? at (0,4)

Solution Steps

To solve this problem, we need to find the critical points of the function \( f(x, y) = x^2 y + y^3 - 48y \) and determine the nature of these critical points (whether they are maxima, minima, or saddle points). Critical points occur where the partial derivatives of the function with respect to \( x \) and \( y \) are both zero.

1. Compute the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).
2. Set these partial derivatives to zero and solve the resulting system of equations to find the critical points.
3. Use the second derivative test to classify the nature of each critical point.

We start with the function \( f(x, y) = x^2 y + y^3 - 48y \). The partial derivatives are calculated as follows: \[ f_x = \frac{\partial f}{\partial x} = 2xy \] \[ f_y = \frac{\partial f}{\partial y} = x^2 + 3y^2 - 48 \]

To find the critical points, we set the partial derivatives equal to zero: \[ 2xy = 0 \] \[ x^2 + 3y^2 - 48 = 0 \] From \( 2xy = 0 \), we have two cases: \( x = 0 \) or \( y = 0 \).

1. If \( x = 0 \): \[ 3y^2 - 48 = 0 \implies y^2 = 16 \implies y = \pm 4 \] This gives us the points \( (0, 4) \) and \( (0, -4) \).

2. If \( y = 0 \): \[ x^2 - 48 = 0 \implies x^2 = 48 \implies x = \pm 4\sqrt{3} \] This gives us the points \( (4\sqrt{3}, 0) \) and \( (-4\sqrt{3}, 0) \).

Thus, the critical points are: \[ (0, -4), (0, 4), (-4\sqrt{3}, 0), (4\sqrt{3}, 0) \]

Next, we use the second derivative test to classify the critical points. The second partial derivatives are: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2y \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 6y \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 2x \]

We compute the determinant \( D \) for each critical point: \[ D = f_{xx} f_{yy} - (f_{xy})^2 \]

1. For \( (0, -4) \): \[ f_{xx} = 2(-4) = -8, \quad f_{yy} = 6(-4) = -24, \quad f_{xy} = 2(0) = 0 \] \[ D = (-8)(-24) - 0^2 = 192 > 0 \quad \text{and} \quad f_{xx} < 0 \implies \text{local maximum} \]

2. For \( (0, 4) \): \[ f_{xx} = 2(4) = 8, \quad f_{yy} = 6(4) = 24, \quad f_{xy} = 2(0) = 0 \] \[ D = (8)(24) - 0^2 = 192 > 0 \quad \text{and} \quad f_{xx} > 0 \implies \text{local minimum} \]

3. For \( (-4\sqrt{3}, 0) \): \[ f_{xx} = 2(0) = 0, \quad f_{yy} = 6(0) = 0, \quad f_{xy} = 2(-4\sqrt{3}) = -8\sqrt{3} \] \[ D = 0 \quad \text{(inconclusive)} \]

4. For \( (4\sqrt{3}, 0) \): \[ f_{xx} = 2(0) = 0, \quad f_{yy} = 6(0) = 0, \quad f_{xy} = 2(4\sqrt{3}) = 8\sqrt{3} \] \[ D = 0 \quad \text{(inconclusive)} \]

Final Answer