Problem 10 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)-x^{2}-2 x y+3 y^{2}-8 y $$

Step-by-Step Solution

Answer

The critical point \((-1, 1)\) is a saddle point.

Step 1: Find the Partial Derivatives

To find the critical points, we need the partial derivatives of the function. The first partial derivative with respect to \(x\) is \( \frac{\partial f}{\partial x} = -2x - 2y \), and with respect to \(y\) is \( \frac{\partial f}{\partial y} = -2x + 6y - 8 \).

Step 2: Set Partial Derivatives to Zero

Set each partial derivative equation to zero to find the critical points. Solving \(-2x - 2y = 0\), we get \(x = -y\). For \(-2x + 6y - 8 = 0\), substitute \(x = -y\) to yield \(-2(-y) + 6y - 8 = 0\). Simplify to \(8y - 8 = 0\), then solve, giving \(y = 1\). Substituting \(y = 1\) into \(x = -y\), we have \(x = -1\).

Step 3: Use the Second Derivative Test

Find the second partial derivatives: \( f_{xx} = -2 \), \( f_{yy} = 6 \), and \( f_{xy} = -2 \). Evaluate the determinant of the Hessian matrix at the critical point \((-1, 1)\). The determinant \(D\) is given by \(D = f_{xx}f_{yy} - (f_{xy})^2 = (-2)(6) - (-2)^2 = -12 - 4 = -16\).

Step 4: Determine the Nature of the Critical Points

With \(D < 0\), the critical point \((-1, 1)\) is a saddle point, indicating it is neither a local maximum nor a local minimum.

Key Concepts

Partial Derivatives

Partial derivatives are used to understand how a function changes as its input variables change. When dealing with multivariable functions, like in our original exercise, we use partial derivatives to find critical points. Critical points occur where the gradient (which consists of all partial derivatives) is zero. For the function \( f(x, y) = x^2 + 2xy - 3y^2 + 8y \), the process begins by finding the partial derivatives with respect to \( x \) and \( y \).

• The derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = -2x - 2y \)
• The derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = -2x + 6y - 8 \)

Setting these equations to zero helps us locate where the function potentially changes behavior: either reaching a peak, a valley, or a saddle.

Second Derivative Test

Once the partial derivatives identify critical points, the second derivative test helps us determine the nature of these points. In single-variable calculus, the second derivative test is straightforward. However, for multivariable functions, this test requires more steps.The second derivatives provide information on the concavity of the function. By calculating the second partial derivatives, we gain insight into how the curvature of the function changes around critical points. In this exercise, we found:

• \( f_{xx} = -2 \)
• \( f_{yy} = 6 \)
• \( f_{xy} = -2 \)

These derivatives are essential to construct the Hessian matrix, which is used in the next step to classify the critical points.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives. It plays a crucial role in determining whether a critical point is a local maximum, a local minimum, or a saddle point.The Hessian matrix for our function \( f(x, y) \) at the critical point \((-1, 1)\) is:\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{xy} & f_{yy}\end{bmatrix} = \begin{bmatrix}-2 & -2 \-2 & 6\end{bmatrix}\]The determinant of the Hessian matrix, \( D \), is calculated by:\[D = f_{xx} f_{yy} - (f_{xy})^2 = (-2)(6) - (-2)^2 = -16\]The value of \( D \) indicates the nature of the critical point.

Local Maximum

Determining whether a critical point is a local maximum involves using the Hessian determinant, \( D \), and the value of \( f_{xx} \). For local maxima, the following conditions should hold:

• \( D > 0 \)
• \( f_{xx} < 0 \)

In our exercise, we found that \( D = -16 \), which is less than zero. Therefore, the critical point \((-1, 1)\) does not qualify as a local maximum. This result suggests that the function does not have a peak at this point.

Local Minimum

A local minimum occurs when the determinant of the Hessian matrix and the sign of \( f_{xx} \) meet specific criteria:

• \( D > 0 \)
• \( f_{xx} > 0 \)

For the given critical point \((-1, 1)\), the determinant \( D \) was found to be \(-16\), indicating a negative value. Since \( D < 0 \), we conclude that the critical point is not a local minimum. In fact, a negative \( D \) tells us the critical point is a saddle point, meaning it is neither a peak nor a valley but rather a transition area in the surface of the graph of the function.