Problem 316 - Calculus Volume 3

Question

For the following exercises, find the critical points of the function by using algebraic techniques (completing the square) or by examining the form of the equation. Verify your results using the partial derivatives test. $$f(x, y)=x^{2}+y^{2}+2 x-6 y+6$$

Step-by-Step Solution

Answer

The critical point is (-1, 3), a local minimum.

Step 1: Write Down the Function

To find the critical points of the function, we start by writing down the given function: \[ f(x, y) = x^2 + y^2 + 2x - 6y + 6 \]

Step 2: Find the Partial Derivatives

Calculate the partial derivatives of the function with respect to \(x\) and \(y\):- The partial derivative with respect to \(x\) is: \[ f_x = \frac{\partial}{\partial x}(x^2 + y^2 + 2x - 6y + 6) = 2x + 2 \]- The partial derivative with respect to \(y\) is: \[ f_y = \frac{\partial}{\partial y}(x^2 + y^2 + 2x - 6y + 6) = 2y - 6 \]

Step 3: Set the Partial Derivatives to Zero

To find the critical points, set each of the partial derivatives to zero and solve:- For \(f_x = 0\): \[ 2x + 2 = 0 \] Solving this gives: \( x = -1 \).- For \(f_y = 0\): \[ 2y - 6 = 0 \] Solving this gives: \( y = 3 \).

Step 4: Determine the Critical Point

Combine the solutions to determine the critical point:The critical point is \((-1, 3)\).

Step 5: Verify Using Second Partial Derivative Test

Utilize the second partial derivative test to verify the nature of the critical point:- Compute second partial derivatives: \[ f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \] \[ f_{yy} = \frac{\partial^2 f}{\partial y^2} = 2 \] \[ f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \]- Calculate the determinant of the Hessian matrix: \[ D = f_{xx}f_{yy} - (f_{xy})^2 = 2 \cdot 2 - 0^2 = 4 \] Since \(D > 0\) and \(f_{xx} > 0\), the critical point \((-1, 3)\) is a local minimum.

Key Concepts

Partial Derivatives

Partial derivatives are a fundamental concept in multivariable calculus used to analyze functions with more than one variable. Think of them as the building blocks to understand how a function changes with respect to one of its variables while keeping others constant.

• For a function \(f(x, y)\), the partial derivative with respect to \(x\), denoted as \(f_x\) or \(\frac{\partial f}{\partial x}\), measures the rate of change of the function as \(x\) changes, while \(y\) is held constant.


• Similarly, the partial derivative with respect to \(y\), denoted as \(f_y\), indicates how the function changes as \(y\) varies, with \(x\) fixed.

In the exercise, we calculated these derivatives for the function \(f(x, y) = x^2 + y^2 + 2x - 6y + 6\) to be \(f_x = 2x + 2\) and \(f_y = 2y - 6\). Setting these equal to zero helps locate potential critical points where the function does not change with small alterations in \(x\) or \(y\).

Second Partial Derivative Test

The Second Partial Derivative Test is a tool to classify critical points found using partial derivatives. It tells us whether a critical point is a local minimum, local maximum, or saddle point.
When you have a function \(f(x, y)\):

• Calculate the second partial derivatives: \(f_{xx}\), \(f_{yy}\), and \(f_{xy}\).


• Evaluate the determinant of the Hessian matrix, which is given by \(D = f_{xx}f_{yy} - (f_{xy})^2\).

• If \(D > 0\) and \(f_{xx} > 0\), the critical point is a local minimum.
• If \(D > 0\) and \(f_{xx} < 0\), it is a local maximum.
• If \(D < 0\), the critical point is a saddle point.

In our exercise, by computing these second derivatives, we verified that the critical point \((-1, 3)\) is indeed a local minimum by confirming \(D = 4 > 0\) and \(f_{xx} = 2 > 0\).

Hessian Matrix

The Hessian Matrix is a square matrix of second-order partial derivatives of a function. It is crucial when applying the Second Partial Derivative Test.
For a function \(f(x, y)\), the Hessian matrix \(H\) is:
\[H = \begin{bmatrix}f_{xx} & f_{xy} \f_{xy} & f_{yy}\end{bmatrix}\]

• \(f_{xx}\) is the second partial derivative with respect to \(x\),


• \(f_{yy}\) is the second partial derivative with respect to \(y\), and


• \(f_{xy}\) is the mixed partial derivative.

By analyzing the Hessian matrix, especially its determinant \(D\), we decide the nature of the critical point. In this exercise, the Hessian matrix is:
\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}\]
The determinant \(D = 4\) confirms the point \((-1, 3)\) as a local minimum.

Local Minimum

A local minimum for a function \(f(x, y)\) is a point where the function has the lowest value compared to nearby points. This means the function dips down at this point, similar to a valley.
At a local minimum:

• The function value at the point is less than or equal to the function values at nearby points.


• In the context of derivative tests, both the first and second derivative tests confirm the nature of the function at that point.

In this exercise, after setting partial derivatives to zero and applying the Second Partial Derivative Test, we found that \((-1, 3)\) is a local minimum for \(f(x, y) = x^2 + y^2 + 2x - 6y + 6\). The confirmation was done through the positive determinant of the Hessian matrix and the positive second partial derivative with respect to \(x\). These calculations verify that at this point, the function exhibits a "valley" behavior, making it a local minimum.