Problem 11 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)=x^{3}-3 x+y^{3}-3 y $$

Step-by-Step Solution

Answer

Local minimum at (1, 1), local maximum at (-1, -1), saddle points at (-1, 1) and (1, -1).

Step 1: Find the first partial derivatives

To find the critical points, we need to find where the partial derivatives equal zero. Calculate the partial derivatives with respect to each variable:For \( x \):\[ f_x = \frac{\partial}{\partial x}(x^3 - 3x + y^3 - 3y) = 3x^2 - 3 \]For \( y \):\[ f_y = \frac{\partial}{\partial y}(x^3 - 3x + y^3 - 3y) = 3y^2 - 3 \]

Step 2: Solve for critical points

Set each partial derivative equal to zero and solve for the variables:\( f_x = 0 \) gives:\[ 3x^2 - 3 = 0 \]\[ x^2 = 1 \]\[ x = \pm 1 \]\( f_y = 0 \) gives:\[ 3y^2 - 3 = 0 \]\[ y^2 = 1 \]\[ y = \pm 1 \]Thus, the critical points are \((1, 1), (-1, 1), (1, -1), (-1, -1)\).

Step 3: Use the Second Derivative Test

To determine the nature of each critical point, use the second partial derivatives to calculate the Hessian matrix. Calculate the second derivatives:\( f_{xx} = \frac{\partial^2}{\partial x^2} f = 6x \)\( f_{yy} = \frac{\partial^2}{\partial y^2} f = 6y \)\( f_{xy} = f_{yx} = \frac{\partial^2}{\partial x \partial y} f = 0 \)The Hessian matrix \( H \) is:\[H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} = \begin{bmatrix} 6x & 0 \ 0 & 6y \end{bmatrix} \]

Step 4: Evaluate Hessian determinant at critical points

Evaluate the determinant of the Hessian \( D = f_{xx} f_{yy} - (f_{xy})^2 \) at each critical point:- At \( (1, 1) \): \[ D = 6 \cdot 1 \cdot 6 \cdot 1 - 0^2 = 36 \]- At \( (-1, 1) \): \[ D = 6 \cdot (-1) \cdot 6 \cdot 1 - 0^2 = -36 \]- At \( (1, -1) \): \[ D = 6 \cdot 1 \cdot 6 \cdot (-1) - 0^2 = -36 \]- At \( (-1, -1) \): \[ D = 6 \cdot (-1) \cdot 6 \cdot (-1) - 0^2 = 36 \]

Step 5: Determine the nature using Hessian determinant

Use the second derivative test:- If \( D > 0 \) and \( f_{xx} > 0 \), there is a local minimum.- If \( D > 0 \) and \( f_{xx} < 0 \), there is a local maximum.- If \( D < 0 \), it is a saddle point.For the points:- At \( (1, 1) \): \( D = 36 > 0 \) and \( f_{xx} = 6 \cdot 1 > 0 \), so there is a local minimum.- At \( (-1, 1) \) and \( (1, -1) \): \( D = -36 < 0 \), so they are saddle points.- At \( (-1, -1) \): \( D = 36 > 0 \) and \( f_{xx} = 6 \cdot (-1) < 0 \), so there is a local maximum.

Key Concepts

Partial Derivatives

Partial derivatives are essential when dealing with functions of more than one variable. They represent the rate of change of the function with respect to one variable while keeping others constant. Imagine analyzing how a landscape changes as you move in just one direction at a time, like moving along the x-axis first, and then the y-axis.

• In our exercise: We computed partial derivatives \( f_x \) and \( f_y \) with respect to \( x \) and \( y \), respectively:

- \( f_x = 3x^2 - 3 \), calculating it shows how changes in \( x \) affect the function.
- \( f_y = 3y^2 - 3 \), similarly indicates how changes in \( y \) affect the function.

To locate critical points, we set these partial derivatives to zero, as zeroes point to potential maxima, minima, or saddle points.

Second Derivative Test

The Second Derivative Test helps us determine the nature of critical points identified by our first derivatives. It tells us if each point is a local maximum, a local minimum, or a saddle point.

• In this test, we calculate second derivatives and use them to assess the curvature around each critical point.

To apply this test, compute the second derivatives such as \( f_{xx}, f_{yy}, \) and cross-derivatives like \( f_{xy} \). This information is crucial to understand how the function behaves near the critical point.

The results let us check the concavity; when \( D = f_{xx} \cdot f_{yy} - (f_{xy})^2 \) is positive, small neighborhoods are either bowl-like or cap-like, corresponding to minima or maxima. If \( D \) is negative, the point is a saddle.

Hessian Matrix

The Hessian matrix is pivotal in multivariable calculus. It is a square matrix of second-order partial derivatives of a scalar-valued function. Think of the Hessian as a tool to encapsulate the second derivative information concisely.

• The structure is typically \( H = \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \).

For our exercise, the Hessian is built using:
- \( f_{xx} = 6x \)
- \( f_{yy} = 6y \)
- \( f_{xy} = f_{yx} = 0 \)

The determinant of this matrix, \( D = f_{xx} f_{yy} - (f_{xy})^2 \), helps define the nature of critical points, acting as a compass for understanding the geometry of the surface near each point.

Local Maximum

A local maximum is where the function value is greater than the values at nearby points. Picture standing on a peak, where every direction you step down from it leads to lower ground.

• A point is a local maximum when \( D > 0 \) and \( f_{xx} < 0 \).

In our function, the point \((-1, -1)\) was a local maximum because:
- The Hessian determinant \( D = 36 \) was positive.
- \( f_{xx} = 6 \cdot (-1) = -6 \) is negative, indicating a "hilltop" locally at that point.

This allows us to classify such points confidently and distinguish them from other types of critical points.

Local Minimum

At a local minimum, the function value is less than those nearby. Imagine a valley where stepping any direction leads to higher ground.

• Occurs when \( D > 0 \) and \( f_{xx} > 0 \).

For our function, the critical point \((1, 1)\) was identified as a local minimum due to:
- Hessian determinant \( D = 36 \), indicating a bowl-like shape.
- \( f_{xx} = 6 \cdot 1 = 6 \) being positive, confirming a "bowl-shaped" minimum.

Understanding local minima is crucial in both theory and applications where minimization of a function implies optimal solutions.

Saddle Point

A saddle point is neither a peak nor a valley. It resembles a saddle where directions can either ascend or descend. Think of a mountain pass or a horse saddle, with both upward and downward slopes.

• Determined by \( D < 0 \), indicating curvature change direction-wise.

In our example, saddle points were found at \((1, -1)\) and \((-1, 1)\) due to:
- Negative determinant values, such as \( D = -36 \) at these points.

Such points highlight instability in optimization problems because they don't offer stable extrema like maxima or minima do.