Problem 1 - Calculus with Concepts in Calculus

Question

Find all critical points. Determine whether each critical point yields a relative maximum value, a relative minimum value, or a saddle point. $$ f(x, y)=x^{2}+2 y^{2}-6 x+8 y-1 $$

Step-by-Step Solution

Answer

The function has one critical point at (3, -2), which is a relative minimum.

Step 1: Find the First Partial Derivatives

To find critical points, compute the first partial derivatives of the function with respect to both variables:The partial derivative with respect to \(x\) is:\[ f_x = \frac{\partial}{\partial x}(x^2 + 2y^2 - 6x + 8y - 1) = 2x - 6 \]The partial derivative with respect to \(y\) is:\[ f_y = \frac{\partial}{\partial y}(x^2 + 2y^2 - 6x + 8y - 1) = 4y + 8 \]

Step 2: Set Partial Derivatives Equal to Zero

Set the first partial derivatives equal to zero to find the critical points:\[ 2x - 6 = 0 \quad \Rightarrow \quad x = 3 \]\[ 4y + 8 = 0 \quad \Rightarrow \quad y = -2 \]Thus, the critical point is \((x, y) = (3, -2)\).

Step 3: Find the Second Partial Derivatives

Now, find the second partial derivatives to use in the second derivative test:\[ f_{xx} = \frac{\partial^2}{\partial x^2}(x^2 + 2y^2 - 6x + 8y - 1) = 2 \]\[ f_{yy} = \frac{\partial^2}{\partial y^2}(x^2 + 2y^2 - 6x + 8y - 1) = 4 \]\[ f_{xy} = \frac{\partial^2}{\partial x \partial y}(x^2 + 2y^2 - 6x + 8y - 1) = 0 \]

Step 4: Apply the Second Derivative Test

Use the second derivative test to classify the critical point:Compute the Hessian determinant:\[ H = f_{xx}f_{yy} - (f_{xy})^2 = (2)(4) - (0)^2 = 8 \]Since \(H > 0\) and \(f_{xx} > 0\), the function has a relative minimum at the critical point \((3, -2)\).

Key Concepts

Partial Derivatives

In calculus, partial derivatives are crucial for finding the critical points of multivariable functions. They help us see how a function changes as we tweak each variable one at a time. For a function like \( f(x, y) = x^2 + 2y^2 - 6x + 8y - 1 \), we focus on each variable separately. For example:

• The partial derivative with respect to \(x\) is found by treating \(y\) as a constant and differentiating the expression for \(x\). For our function, this gives \( f_x = 2x - 6 \).
• The partial derivative with respect to \(y\) treats \(x\) as constant, resulting in \( f_y = 4y + 8 \).

To find critical points, we set these derivatives to zero. This tells us where the slope in any given direction is zero, which often indicates a peak, valley, or saddle point in the function's graph. In our exercise, solving \( 2x - 6 = 0 \) and \( 4y + 8 = 0 \) gives the critical point \((3, -2)\). This is the starting point for further analyzing the behavior of the function at this point.

Second Derivative Test

The second derivative test is an extension of finding critical points. Once we have these points, we use second derivatives to determine the nature of these points. The test involves calculating the second partial derivatives:

• For \( f(x, y) = x^2 + 2y^2 - 6x + 8y - 1 \), the second partial derivative with respect to \(x\) is \( f_{xx} = 2 \), showing how the function curves along the \(x\) direction.
• Similarly, \( f_{yy} = 4 \) shows curvature along the \(y\) direction.
• Mixed partial derivative, \( f_{xy} = 0 \), indicates there is no twist in the surface at this point in terms of \(x\) and \(y\).

Using these values, we apply the second derivative test. Calculate the Hessian determinant: \( H = f_{xx}f_{yy} - (f_{xy})^2 \). If \(H > 0\) and \(f_{xx} > 0\), it indicates a local minimum, while \(f_{xx} < 0\) indicates a local maximum. If \(H < 0\), it suggests a saddle point.

Hessian Determinant

The Hessian determinant is a key player in understanding the nature of critical points. It combines information from second partial derivatives to give a single value that helps determine whether a point is a minimum, maximum, or saddle point.

• For the function \( f(x, y) = x^2 + 2y^2 - 6x + 8y - 1 \), the Hessian matrix is formed using second partials: \( \begin{bmatrix} f_{xx} & f_{xy} \ f_{yx} & f_{yy} \end{bmatrix} \).
• Computing the determinant, \( H = f_{xx}f_{yy} - (f_{xy})^2 \), provides a numerical criterion: \(H = (2)(4) - (0)^2 = 8 \).

The sign of \(H\) reveals the curvature behavior of the function at the critical point. In our example, since \(H > 0\) and the second derivative \(f_{xx}\) is also positive, the critical point at \((3, -2)\) is a relative minimum. This is valuable because it tells us the function dips to its lowest point in this small area, helping us predict its behavior without drawing a graph.