Question

Find and evaluate the maxima, minima and saddle points of the function $$ f(x, y)=x y\left(x^{2}+y^{2}-1\right) $$

Short answer

The function has saddle points at \((0, \text{±1})\) and \((\text{±1}, 0)\). The point \((0,0)\) requires further analysis.

Step-by-step solution

Find the First Order Partial Derivatives

Compute the first partial derivatives of the function with respect to both variables. For \( f(x, y) = xy(x^2 + y^2 - 1) \), the partial derivatives are: \( f_x = \frac{\partial f}{\partial x} = y(3x^2 + y^2 - 1) \) \( f_y = \frac{\u partial f}{\u partial y} = x(x^2 + 3y^2 - 1) \)

Set the Partial Derivatives to Zero

Solve the equations \( f_x = 0 \) and \( f_y = 0 \) to find the critical points. \( y(3x^2 + y^2 - 1) = 0 \) and \( x(x^2 + 3y^2 - 1) = 0 \).

Solve the System of Equations

Analyze the system of equations from the previous step: 1. \(y = 0\) 2. \(3x^2 + y^2 - 1 = 0\) 3. \(x = 0\) 4. \(x^2 + 3y^2 - 1 = 0\). Find all possible critical points. They are \((0, \text{±1}), (\text{±1}, 0), (0,0)\).

Second Order Partial Derivatives

Compute the second partial derivatives: \( f_{xx} = \frac{\u partial^2 f}{\u partial x^2} = 6xy \) \( f_{yy} = \frac{\u partial^2 f}{\u partial y^2} = 6xy \) \( f_{xy} = \frac{\u partial^2 f}{\u partial x \u partial y} = 3x^2 + 3y^2 - 1 \)

Evaluate the Determinant of the Hessian Matrix

Use the second derivatives to form the Hessian matrix and evaluate its determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) at each critical point: 1. \( (0, \text{±1}): D = (0)(0) - (-1)^2 = -1 \) 2. \( (\text{±1}, 0): D = (0)(0) - (-1)^2 = -1 \) 3. \( (0,0): D = 0 \)

Classify the Critical Points

Determine the nature of each critical point based on the value of \( D \): 1. \((0, \text{±1})\) and \((\text{±1}, 0)\): \(D < 0\), so these are saddle points. 2. \((0, 0)\): \(D = 0\), the test is inconclusive.

Key concepts

Maxima

In mathematics, a **maximum** identifies the highest value of a function within a particular region.
When studying functions with two variables, such as our given function \( f(x, y) = xy(x^2 + y^2 - 1) \), we look for points where the function reaches its highest value.
These points are determined by analyzing the partial derivatives and the Hessian matrix. If both the first partial derivatives at a point are zero, and the Hessian matrix indicates a positive definiteness, then we have a local maximum at that point.
However, in the given step-by-step solution, no critical points were classified as maxima based on their Hessian matrices.

Minima

Just like maxima, a **minimum** identifies the lowest value of a function in a region. For a function \( f(x, y) \), the minima can be found by setting the first partial derivatives to zero and examining the Hessian matrix.
Specifically, to identify a minimum, we require the determinant of the Hessian to be positive and the second order partial derivatives should indicate a node with positive concavity. This was analyzed in our solution, but we did not find any values indicating a minimum. All critical points found did not satisfy the required conditions for minima.

Saddle Points

A **saddle point** is where the function behaves like a saddle. It is a point where the function is not giving a maximum or a minimum. Instead, the function curves upward in one direction and downward in another.
For our function \( f(x, y) = xy(x^2 + y^2 - 1) \), we found critical points by solving the system of first-order partial derivatives. After checking these points using the Hessian determinant \( D \), we observed that \( D \) was negative at the critical points \((0, \text{±1})\) and \((\text{±1}, 0)\). This indicates that these points are saddle points. Saddle points are crucial in understanding the local curvature of a function.

Partial Derivatives

A **partial derivative** represents how a function changes as one variable changes while keeping others constant. It is a foundational concept in multivariable calculus.
For our function \( f(x, y) \), the first-order partial derivatives are:
\( f_x = \frac{\partial f}{\partial x} = y(3x^2 + y^2 - 1) \)
\( f_y = \frac{\partial f}{\partial y} = x(x^2 + 3y^2 - 1) \).
Setting these derivatives to zero allows us to find the critical points. In our case, the critical points found were \((0, \text{±1})\), \((\text{±1}, 0)\), and \((0, 0)\). These partial derivatives help in finding where the function is stationary.

Hessian Matrix

The **Hessian matrix** is a square matrix of second-order partial derivatives of a function. It provides critical information about the local curvature of that function. For a function \( f(x, y) \), it is defined as:
\ H = \begin{pmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{pmatrix} \
For our specific function, the second-order partial derivatives are:
\( f_{xx} = 6xy \)
\( f_{yy} = 6xy \)
\( f_{xy} = 3x^2 + 3y^2 - 1 \).
The determinant of the Hessian matrix \( D \) at each critical point helps classify them. If \( D > 0 \) and \( f_{xx} > 0 \), we have a local minimum; if \( D > 0 \) and \( f_{xx} < 0 \), we have a maximum. If \( D < 0 \), it is a saddle point. In our solution, \( D \) was negative or zero, suggesting saddle points or inconclusive results for some points.