Problem 12 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, or neither. $$ f(x, y)=x^{3}+y^{2}-3 x^{2}+10 y+6 $$

Step-by-Step Solution

Answer

Critical points are (0, -5) (saddle point) and (2, -5) (local minimum).

Step 1: Partial Derivatives

To find critical points, we need to calculate the partial derivatives of the function with respect to each variable. For \(f(x, y) = x^3 + y^2 - 3x^2 + 10y + 6\), the partial derivatives are: \[ f_x = \frac{\partial f}{\partial x} = 3x^2 - 6x \] \[ f_y = \frac{\partial f}{\partial y} = 2y + 10 \]

Step 2: Solving for Critical Points

Set the partial derivatives equal to zero to find critical points:1. \(3x^2 - 6x = 0\)2. \(2y + 10 = 0\)Solving these equations:- From \(3x^2 - 6x = 0\), factoring gives \(x(3x - 6) = 0\), so \(x = 0\) or \(x = 2\).- From \(2y + 10 = 0\), we get \(y = -5\).Thus, the critical points are \((0, -5)\) and \((2, -5)\).

Step 3: Second Derivative Test

We use the second derivative test for functions of two variables. First, find the second partial derivatives:\[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 6x - 6 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \]The Hessian determinant \(H\) is given by:\[ H(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (6x - 6)(2) - 0 = 12x - 12 \]

Step 4: Classification of Critical Points

Evaluate the Hessian at each critical point:1. \((0, -5)\): - \(H(0, -5) = 12(0) - 12 = -12\) - Since \(H < 0\), this point is a saddle point (neither a maximum nor a minimum).2. \((2, -5)\): - \(H(2, -5) = 12(2) - 12 = 24 - 12 = 12\) - Since \(H > 0\) and \(f_{xx} > 0\) (\(6(2) - 6 = 6\)), this is a local minimum.

Key Concepts

Partial Derivatives

In the realm of multivariable calculus, partial derivatives play a crucial role in analyzing functions with more than one variable. When examining a function like \( f(x, y) = x^3 + y^2 - 3x^2 + 10y + 6 \), partial derivatives help us understand how the function changes along each axis independently.

• The partial derivative with respect to \( x \), denoted as \( f_x \), provides insights on how \( f \) changes as \( x \) varies, holding \( y \) constant.
• The partial derivative with respect to \( y \), denoted as \( f_y \), describes the rate of change of \( f \) as \( y \) changes, with \( x \) held constant.

For the given function, you calculate:

• \( f_x = 3x^2 - 6x \)
• \( f_y = 2y + 10 \)

These derivatives are essential for finding critical points, where the function's rate of change could signify key features like peaks, troughs, or inflection points. By setting these equal to zero, we can pinpoint the locations of potential critical points.

Second Derivative Test

The Second Derivative Test is a valuable method for classifying critical points found through partial derivatives. After pinpointing these points, we need this test to determine whether each point is a local maximum, local minimum, or neither.
This process involves:

• Calculating the second partial derivatives, such as \( f_{xx} \), \( f_{yy} \), and the cross-derivative \( f_{xy} \).
• Using these derivatives to construct the Hessian determinant, \( H(x, y) \).

The Second Derivative Test provides a systematic approach to evaluating the nature of each critical point, based on the Hessian and the second derivatives.

Hessian Determinant

The Hessian determinant is a crucial element in the Second Derivative Test. It is a finite value calculated from the second partial derivatives and is used to classify the critical points.
To form the Hessian for our function \( f(x,y) \):

• Find \( f_{xx} = 6x - 6 \)
• \( f_{yy} = 2 \)
• \( f_{xy} = 0 \)

The Hessian determinant \( H(x, y) \) is computed as:
\[ H(x, y) = f_{xx}f_{yy} - (f_{xy})^2 = (6x - 6)(2) - 0 = 12x - 12 \]
The value of the Hessian influences whether a critical point is a local maximum, minimum, or neither. Positive and negative values each guide us towards different classifications.

Local Maximum

A local maximum occurs at a critical point where the function has a peak. It is defined as a point where the function values are larger than the surrounding points in a neighborhood.
In the context of our problem, using the second derivative test:

• If the Hessian determinant \( H(x, y) > 0 \) and the second partial derivative \( f_{xx} < 0 \), the critical point is classified as a local maximum.

If a point had these conditions, it would mean the function cabin at that point from all directions, creating a peak.

Local Minimum

A local minimum at a critical point exists when a function takes on a smaller value than at all nearby points. For our problem:

• The Hessian determinant \( H(x, y) > 0 \).
• The second partial derivative \( f_{xx} > 0 \).

This indicates a local minimum. In our exercise, the critical point \((2, -5)\) was found to be a local minimum due to \( H(2, -5) > 0 \) and \( f_{xx} > 0 \).

Such a scenario means the function has a trough, dipping down at this point compared to its surroundings. Understanding these conditions aids in graphically perceiving the behavior of multivariable functions.