Problem 7 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)=x^{2}+x y+3 y $$

Step-by-Step Solution

Answer

The critical point \((-3, 6)\) is a saddle point (neither max nor min).

Step 1: Find the Partial Derivatives

To find the critical points, we'll start by finding the first partial derivatives of the function \( f(x, y) = x^2 + xy + 3y \). Compute \( f_x \) and \( f_y \):1. \( f_x = \frac{\partial}{\partial x}(x^2 + xy + 3y) = 2x + y \).2. \( f_y = \frac{\partial}{\partial y}(x^2 + xy + 3y) = x + 3 \).

Step 2: Set the Partial Derivatives to Zero

To locate the critical points, set the partial derivatives equal to zero and solve for \( x \) and \( y \):1. \( 2x + y = 0 \).2. \( x + 3 = 0 \).Solving \( x + 3 = 0 \) gives \( x = -3 \). Substitute \( x = -3 \) into the first equation:\[ 2(-3) + y = 0 \Rightarrow y = 6 \].Thus, the critical point is \( (-3, 6) \).

Step 3: Calculate the Second Partial Derivatives

Next, we need to determine the nature of the critical point using the second derivatives. Compute the second partial derivatives:1. \( f_{xx} = \frac{\partial}{\partial x}(2x + y) = 2 \).2. \( f_{yy} = \frac{\partial}{\partial y}(x + 3) = 0 \).3. \( f_{xy} = \frac{\partial}{\partial y}(2x + y) = 1 \).

Step 4: Determine the Nature of Each Critical Point

To analyze the critical point \( (-3, 6) \), use the second derivative test via the Hessian determinant \( D \):\[ D = f_{xx}f_{yy} - (f_{xy})^2 = (2)(0) - (1)^2 = -1 \].If \( D < 0 \), then the critical point is a saddle point (neither a local max nor a local min). Thus, \( (-3, 6) \) is a saddle point.

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, partial derivatives are essential tools. They help in understanding how the function changes with respect to each variable individually. In our exercise, we started by calculating the partial derivatives of the function \( f(x, y) = x^2 + xy + 3y \). This involves differentiating the function with respect to each variable while treating other variables as constants.
For \( f(x, y) \), the partial derivative with respect to \( x \), denoted as \( f_x \), is calculated by treating \( y \) as a constant. Similarly, for the partial derivative with respect to \( y \), \( f_y \), \( x \) is treated as a constant.

• \( f_x = 2x + y \)
• \( f_y = x + 3 \)

These results help identify critical points by locating where both \( f_x \) and \( f_y \) equal zero.

Local Maximum

A local maximum refers to a point where the function reaches a peak value in a specific region or neighborhood. This means that in close proximity to this point, there is no higher function value. To determine if a critical point is a local maximum, one needs to examine the second derivatives. However, in our exercise, the second derivative test showed that the critical point was neither a maximum nor a minimum.
Typically, if a point is a local maximum, this will be confirmed if the Hessian determinant \( D > 0 \) and the second partial derivative with respect to \( x \) is negative (\( f_{xx} < 0 \)), indicating a concave down curve at that point.

Local Minimum

A local minimum indicates a point on the function where it attains a minimum value in its immediate area. To find such points, we again refer to the second derivatives. In our exercise, we assessed whether the only critical point \((-3, 6)\) was a local minimum using the second derivative test.
For a critical point to be a local minimum:

• The Hessian determinant \( D \) must be greater than zero \( (D > 0) \).
• The second partial derivative with respect to \( x \) (\( f_{xx} \)) should be positive (\( f_{xx} > 0 \)), suggesting that the curve is concave up at the point.

In our case, \( D = -1 \), indicating neither a local maximum nor a local minimum.

Second Derivative Test

The second derivative test is a valuable tool in determining the nature of a critical point found from a multivariable function. After identifying critical points using partial derivatives, the next step involves computing the second partial derivatives. These include \( f_{xx} \), \( f_{yy} \), and \( f_{xy} \), and they help form the Hessian determinant \( D \) given by:\[D = f_{xx}f_{yy} - (f_{xy})^2\]The sign of the determinant \( D \) is crucial:

• If \( D > 0 \) and \( f_{xx} > 0 \), the critical point is a local minimum.
• If \( D > 0 \) and \( f_{xx} < 0 \), the critical point is a local maximum.
• If \( D < 0 \), the critical point is a saddle point, indicating neither a local maximum nor minimum.

With \( D = -1 \) in our exercise, the point \((-3, 6)\) was identified as a saddle point. The test shows that despite being a critical point, it doesn’t lead to any extremum within its vicinity.