Problem 318 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=-x^{3}+4 x y-2 y^{2}+1 $$

Step-by-Step Solution

Answer

(0,0) is a saddle point; \( \left(\frac{4}{3}, \frac{4}{3}\right) \) is a local maximum.

Step 1: Find the First Partial Derivatives

To find the critical points, we first need the first partial derivatives of the function \( f(x, y) = -x^3 + 4xy - 2y^2 + 1 \). Calculate the partial derivative with respect to \( x \):\[ f_x = \frac{\partial}{\partial x}(-x^3 + 4xy - 2y^2 + 1) = -3x^2 + 4y \]Now, find the partial derivative with respect to \( y \):\[ f_y = \frac{\partial}{\partial y}(-x^3 + 4xy - 2y^2 + 1) = 4x - 4y \]

Step 2: Solve for Critical Points

Set the first partial derivatives equal to zero to find the critical points:\[ -3x^2 + 4y = 0 \] \[ 4x - 4y = 0 \]From the second equation, solve for \( y \):\[ y = x \]Substitute \( y = x \) into the first equation:\[ -3x^2 + 4x = 0 \]Factor the equation:\[ x(-3x + 4) = 0 \]Thus, \( x = 0 \) or \( x = \frac{4}{3} \), giving the critical points \((0, 0)\) and \( \left( \frac{4}{3}, \frac{4}{3} \right) \).

Step 3: Determine Second Partial Derivatives

Calculate the second partial derivatives needed for the second derivative test:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(-x^3 + 4xy - 2y^2 + 1) = -6x \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(-x^3 + 4xy - 2y^2 + 1) = -4 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(-x^3 + 4xy - 2y^2 + 1) = 4 \]

Step 4: Evaluate the Second Derivative Test

The second derivative test uses the determinant of the Hessian matrix, \( D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 \).First, evaluate at the critical point \((0, 0)\):\[ D(0,0) = (-6 \cdot 0)(-4) - (4)^2 = -16 \]Since \( D(0,0) < 0 \), this point is a saddle point.Next, evaluate at \( \left( \frac{4}{3}, \frac{4}{3} \right) \):\[ f_{xx}\left(\frac{4}{3}, \frac{4}{3}\right) = -6 \cdot \frac{4}{3} = -8 \]\[ D\left(\frac{4}{3}, \frac{4}{3}\right) = (-8)(-4) - (4)^2 = 32 - 16 = 16 \]Since \( D\left(\frac{4}{3}, \frac{4}{3}\right) > 0 \) and \( f_{xx}\left(\frac{4}{3}, \frac{4}{3}\right) < 0 \), this point is a local maximum.

Key Concepts

Critical Points

In calculus, critical points of a function are the values in the domain where the first partial derivatives are zero or undefined. For a two-variable function like \( f(x, y) \), we first compute \( f_x \) and \( f_y \), which represent the partial derivatives with respect to \( x \) and \( y \) respectively.

To find critical points, set \( f_x = 0 \) and \( f_y = 0 \) and solve this system of equations. These solutions give us specific \( (x, y) \) points, which are known as critical points. These points are important because they potentially indicate locations of maxima, minima, or saddle points in the function.

• For the given function \( f(x, y) = -x^3 + 4xy - 2y^2 + 1 \), the critical points were found to be \((0,0)\) and \(\left(\frac{4}{3}, \frac{4}{3}\right)\).

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus. They measure how a function changes as one of the input variables is infinitesimally changed, holding all other variables constant.

For a function \( f(x, y) \), the partial derivative with respect to \( x \), written as \( f_x \), shows the rate of change of \( f \) as \( x \) changes. Similarly, \( f_y \) reflects change as \( y \) varies.

• In our example, \( f_x = -3x^2 + 4y \) and \( f_y = 4x - 4y \). These derivatives were used to find the critical points by setting them equal to zero.

Understanding partial derivatives is crucial when applying the second derivative test, as they help identify how a surface might increase or decrease in any direction.

Hessian Matrix

The Hessian matrix is a square matrix used in multivariable calculus to analyze local behavior of a function. It contains all the second partial derivatives of the function.

For a two-variable function \( f(x, y) \), the Hessian is:
\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix}\]
The determinant of this matrix, \( D(x, y) = f_{xx}f_{yy} - (f_{xy})^2 \), helps classify critical points.

• If \( D > 0 \) and \( f_{xx} < 0 \), the point is a local maximum.
• If \( D > 0 \) and \( f_{xx} > 0 \), the point is a local minimum.
• If \( D < 0 \), the point is a saddle point.

For the function \( f(x, y) = -x^3 + 4xy - 2y^2 + 1 \), the calculations resulted in a determinant that was negative at \((0,0)\) and positive but with \( f_{xx} < 0 \) at \(\left(\frac{4}{3}, \frac{4}{3}\right)\), indicating a saddle point and a local maximum respectively.

Saddle Point

A saddle point for a function of two variables is a type of critical point where the function does not have a local maximum or minimum. Instead, it forms a 'saddle' shape, resembling a surface that curves upwards in one direction and downwards in another.

These points are detected during the second derivative test when the determinant of the Hessian matrix at that point is negative. This negativity means that the surface near the point is concave in one direction and convex in another.

• In our example, at the critical point \((0, 0)\), the test showed \(D(0,0) = -16 \), indicating a saddle point.

Saddle points are significant in optimization problems because they are not solutions where the function achieves a clean extreme value, yet they are critical for understanding the geometry and behavior of functions.