Problem 316 - Calculus 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 of the function is (-1, 3), which is a local minimum.

Step 1: Find First Partial Derivatives

To identify critical points, calculate the first partial derivatives of the function with respect to both variables. The function is given as \[ f(x, y) = x^2 + y^2 + 2x - 6y + 6. \]The first partial derivative with respect to \( x \) is \[ \frac{\partial f}{\partial x} = 2x + 2. \]The first partial derivative with respect to \( y \) is \[ \frac{\partial f}{\partial y} = 2y - 6. \]

Step 2: Set Partial Derivatives to Zero

To find critical points, set each of the first partial derivatives to zero and solve for the variables. For \( \frac{\partial f}{\partial x} = 0 \):\[ 2x + 2 = 0 \]Solving for \( x \), we get:\[ x = -1. \]For \( \frac{\partial f}{\partial y} = 0 \):\[ 2y - 6 = 0 \]Solving for \( y \), we get:\[ y = 3. \]

Step 3: Verify Critical Point via Second Derivative Test

The critical point found is \((-1, 3)\). Now, use the second derivative test to determine the nature of this critical point. Calculate second partial derivatives:\[ \frac{\partial^2 f}{\partial x^2} = 2, \]\[ \frac{\partial^2 f}{\partial y^2} = 2 \]\[ \frac{\partial^2 f}{\partial x \partial y} = 0. \]The Hessian determinant \( H \) is given by:\[ H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2. \]Substitute the known values:\[ H = 2 \cdot 2 - 0^2 = 4. \]Since \( H > 0 \) and \( \frac{\partial^2 f}{\partial x^2} > 0 \), the critical point \((-1, 3)\) is a local minimum.

Key Concepts

Partial Derivatives

In the realm of multivariable calculus, partial derivatives are vital tools that help assess how a function changes as each variable changes independently. When we have a function like \(f(x, y) = x^2 + y^2 + 2x - 6y + 6,\)we want to know how it behaves as either \(x\) or \(y\) changes.

Partial derivatives involve differentiating the function concerning one variable while keeping other variables constant. In our problem, we computed these by differentiating \(f\) first concerning \(x\) to get\(\frac{\partial f}{\partial x} = 2x + 2\),and second concerning \(y\) to get \(\frac{\partial f}{\partial y} = 2y - 6\).These equations help us find the critical points by setting them equal to zero, solving for \(x\) and \(y\).

Second Derivative Test

The second derivative test for functions with several variables is similar to the concept in single-variable calculus but involves a bit more complexity. After finding a critical point, like \((-1, 3)\),we want to determine its nature-whether it is a local minimum, maximum, or a saddle point. This is done using the second derivative test.

Here, first calculate the second partial derivatives:- \(\frac{\partial^2 f}{\partial x^2} = 2\) - \(\frac{\partial^2 f}{\partial y^2} = 2\)- \(\frac{\partial^2 f}{\partial x \partial y} = 0\)

These will then be used to construct the Hessian determinant, which acts as our primary tool in the second derivative test.

Hessian Determinant

The Hessian determinant is calculated using the second partial derivatives and is a crucial part of the second derivative test. It helps identify the type of critical point. The formula for the Hessian determinant \(H\) is:

• \(H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left( \frac{\partial^2 f}{\partial x \partial y} \right)^2\)

Let's substitute the values obtained:- For our function, \(H = 2 \cdot 2 - 0^2 = 4\)

The sign of the Hessian determinant at the critical point plays a decisive role:- If \(H > 0\) and \(\frac{\partial^2 f}{\partial x^2} > 0\), it indicates a local minimum.This positive result suggests that the function at point \((-1, 3)\)behaves as a bowl opening upward.

Local Minimum

Now that we understand the tools, let's identify what it means for a critical point to be a local minimum. A local minimum is a point where the function has a value lower than any nearby points.

In our specific case, the critical point \((-1, 3)\)is classified as a local minimum because the Hessian determinant is positive and the second derivative \(\frac{\partial^2 f}{\partial x^2}\)confirms the point has a concave up curvature. This suggests that the small neighborhood around \((-1, 3)\)forms a valley or a low point relative to surrounding areas. This is crucial in optimization problems where finding the lowest point is often the objective.