Problem 336 - 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. $$ f(x, y)=-x^{2}-5 y^{2}+10 x-30 y-62 $$

Step-by-Step Solution

Answer

The critical point \((5, -3)\) is a local maximum.

Step 1: Find the First Derivatives

To find critical points, we start by computing the first partial derivatives of the function with respect to both variables (x and y). Calculate \( f_x \) and \( f_y \) as follows: \[ f_x = \frac{\partial f}{\partial x} = -2x + 10 \] \[ f_y = \frac{\partial f}{\partial y} = -10y - 30 \]

Step 2: Set First Derivatives to Zero

Set the first partial derivatives equal to zero to find the critical points.\(-2x + 10 = 0\) simplifies to \(x = 5\) \(-10y - 30 = 0\) simplifies to \(y = -3\) Thus, the critical point is \((5, -3)\).

Step 3: Find the Second Derivatives

Next, calculate the second partial derivatives needed for the second derivative test:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = -2 \]\[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = -10 \]\[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \]

Step 4: Determine Nature of Critical Point

Use the second derivative test:The determinant of the Hessian matrix, \( D \), is calculated as: \[ D = f_{xx}f_{yy} - (f_{xy})^2 = (-2)(-10) - (0)^2 = 20 \]Since \( D > 0 \) and \( f_{xx} < 0 \), the critical point \((5, -3)\) is a local maximum.

Key Concepts

Understanding Critical Points

In the journey to analyze functions, critical points are where the magic begins. These are specific points on a graph where the function's slope is zero or undefined. They are essential because they allow us to potentially find peaks, troughs, or flatter regions of a function.

For any given function, finding critical points involves computing the first derivatives concerning each variable. By setting these derivatives equal to zero, we can solve for the values of the variables that provide these points of interest. In a two-variable scenario, such as with our function, this means finding where both partial derivatives are zero simultaneously.

In the problem at hand, we found the critical point by solving the equations derived from the first partial derivatives. This led us to the coordinates (\(x, y\) = (5, -3)). These coordinates give us the critical point of the function.

Exploring Partial Derivatives

Implied in the name, a partial derivative helps us understand how a function changes as one of its input variables changes, keeping all other variables constant. In simpler terms, it provides the "rate of change" of a function in one direction.

To compute the partial derivatives of our function, \(f(x, y) = -x^2 - 5y^2 + 10x - 30y - 62\), with respect to \(x\) and \(y\), we apply differentiation rules just like in single-variable calculus.

• For \(f_x\), the partial derivative with respect to \(x\), we keep \(y\) constant and differentiate, resulting in \(-2x + 10\).
• For \(f_y\), we differentiate while treating \(x\) as constant, yielding \(-10y - 30\).

By setting these equations to zero, we pin down those values where change momentarily halts, pointing us to potential critical points.

Decoding the Hessian Matrix

The Hessian matrix is like a detective's magnifying glass, helping us distinguish the nature of critical points. It is a square matrix composed of all possible second partial derivatives of a function. Acting as a tool in the second derivative test, it aids in investigating the "curvature" of a function at its critical points.

For our function, the Hessian matrix, \(H\), includes the following elements:

• \(f_{xx} = -2\): The second partial derivative with respect to \(x\).
• \(f_{yy} = -10\): The second partial derivative with respect to \(y\).
• \(f_{xy} = 0\): The mixed second partial derivative.

Using these values, the determinant of the Hessian matrix is calculated as \(D = f_{xx}f_{yy} - (f_{xy})^2 = 20\). This result is central in determining the character of the critical point found.

Identifying a Local Maximum

When it comes to classifying critical points, we're on a quest to understand whether a point is a local maximum, minimum, saddle point, or none of these.

The second derivative test provides a systematic approach using the Hessian matrix. For our function, we determined the determinant of the Hessian matrix \(D = 20\), which is positive. Additionally, we noted that \(f_{xx} = -2\) is negative.

The criteria for identifying a local maximum involves:

• The determinant \(D > 0\)
• The second partial derivative \(f_{xx} < 0\)

With these conditions satisfied, the critical point (5, -3) is confirmed to be a local maximum. This means, within a small neighborhood around this point, the function takes on a maximum value compared to nearby points.