Question

1. consider the function (f) that, for a certain parameter (a in mathbb{r} \setminus \\{0\\}), is defined by

\\f(x, y) = 2ay^2 - ax^2 + 2axy + \frac{y^3}{3} + 4\\

2.0 (a) determine the critical point(s) of (f).
2.0 (b) classify the critical point(s) as local maximum, local minimum or a saddle point according to the values of the parameter (a).

Explanation:

⚡ Using what you learned: Partial Derivatives and Gradients · 🆕 New: Second Derivative Test for Multivariable Functions
Classifying critical points using the Hessian determinant

Step 1: Find the first partial derivatives

To find the critical points, we compute the first-order partial derivatives of \( f(x, y) = 2ay^2 - ax^2 + 2axy + \frac{y^3}{3} + 4 \) with respect to \( x \) and \( y \):

$$ f_x(x, y) = \frac{\partial f}{\partial x} = -2ax + 2ay $$

$$ f_y(x, y) = \frac{\partial f}{\partial y} = 4ay + 2ax + y^2 $$

Step 2: Solve for the critical points

Set both partial derivatives to zero:

1. \( -2ax + 2ay = 0 \implies 2ax = 2ay \)

Since \( a \in \mathbb{R} \setminus \{0\} \), we can divide by \( 2a \):

$$ x = y $$

1. Substitute \( x = y \) into \( f_y(x, y) = 0 \):

$$ 4ay + 2ay + y^2 = 0 \implies y^2 + 6ay = 0 $$

Factor the equation:

$$ y(y + 6a) = 0 $$

This yields two solutions for \( y \):

• \( y_1 = 0 \implies x_1 = 0 \)
• \( y_2 = -6a \implies x_2 = -6a \)

Thus, the critical points are:

$$ (0, 0) \quad \text{and} \quad (-6a, -6a) $$

Step 3: Compute the second-order partial derivatives

To classify these points, we find the second-order partial derivatives to construct the Hessian matrix:

$$ f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2a $$

$$ f_{xy} = \frac{\partial^2 f}{\partial y \partial x} = 2a $$

$$ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 4a + 2y $$

The Hessian determinant \( D(x, y) \) is defined as:

$$ D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (-2a)(4a + 2y) - (2a)^2 = -8a^2 - 4ay - 4a^2 = -12a^2 - 4ay $$

Step 4: Classify the critical point \((0, 0)\)

Evaluate the Hessian determinant at \((0, 0)\):

$$ D(0, 0) = -12a^2 - 4a(0) = -12a^2 $$

Since \( a
eq 0 \), \( a^2 > 0 \), which means:

$$ D(0, 0) = -12a^2 < 0 $$

Because the Hessian determinant is strictly negative, the point \((0, 0)\) is always a saddle point for any \( a \in \mathbb{R} \setminus \{0\} \).

Step 5: Classify the critical point \((-6a, -6a)\)

Evaluate the Hessian determinant at \((-6a, -6a)\):

$$ D(-6a, -6a) = -12a^2 - 4a(-6a) = -12a^2 + 24a^2 = 12a^2 $$

Since \( a
eq 0 \), \( 12a^2 > 0 \), which means \( D(-6a, -6a) > 0 \). This point is a local extremum. We determine its nature using the sign of \( f_{xx} = -2a \):

• If \( a > 0 \), then \( f_{xx} = -2a < 0 \). Since \( D > 0 \) and \( f_{xx} < 0 \), the point \((-6a, -6a)\) is a local maximum.
• If \( a < 0 \), then \( f_{xx} = -2a > 0 \). Since \( D > 0 \) and \( f_{xx} > 0 \), the point \((-6a, -6a)\) is a local minimum.

Answer:

(a) Critical Points:
The critical points of \( f \) are:

$$ (0, 0) \quad \text{and} \quad (-6a, -6a) $$

(b) Classification:

• The critical point \((0, 0)\) is a saddle point for all \( a \in \mathbb{R} \setminus \{0\} \).
• The critical point \((-6a, -6a)\) is:
• A local maximum if \( a > 0 \).
• A local minimum if \( a < 0 \).