Problem 2 - 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. $$ f(x, y)=x^{2}-2 y^{2}-6 x+8 y+3 $$

Step-by-Step Solution

Answer

The critical point is (3, 2), and it is a saddle point.

Step 1: Find the first partial derivatives

To find the critical points of the function \( f(x, y) = x^{2} - 2y^{2} - 6x + 8y + 3 \), we first need to compute the partial derivatives with respect to \( x \) and \( y \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = 2x - 6 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = -4y + 8 \).

Step 2: Set first partial derivatives to zero

To find the critical points, we set the partial derivatives equal to zero: \[ f_x = 2x - 6 = 0 \] and \[ f_y = -4y + 8 = 0 \]. Solving these equations gives us the values of \( x \) and \( y \) for the critical points.

Step 3: Solve for critical points

Solving the equation \( 2x - 6 = 0 \) gives \( x = 3 \). Solving the equation \( -4y + 8 = 0 \) gives \( y = 2 \). Thus, the critical point is \( (3, 2) \).

Step 4: Find the second partial derivatives

Next, we find the second partial derivatives to apply the second derivative test. The second partial derivatives are: \( f_{xx} = \frac{\partial^2 f}{\partial x^2} = 2 \), \( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -4 \), and \( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = 0 \).

Step 5: Apply the second derivative test

Using the second derivative test, calculate the discriminant \[ D = f_{xx}f_{yy} - (f_{xy})^2 \]. Substituting the values, we get \[ D = (2)(-4) - (0)^2 = -8 \]. Since \( D < 0 \), the point \( (3, 2) \) is a saddle point.

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are a way to assess how a function changes as one of its variables is altered while keeping the others constant. For instance, when examining a function of two variables, like our exercise's function \( f(x, y) = x^2 - 2y^2 - 6x + 8y + 3 \), we have two partial derivatives: one with respect to \( x \) and another with respect to \( y \).

For each partial derivative:

• The partial derivative with respect to \( x \), denoted as \( f_x \), is calculated by treating \( y \) as a constant. Here, \( f_x = 2x - 6 \).
• The partial derivative with respect to \( y \), denoted as \( f_y \), is calculated by treating \( x \) as a constant. For our exercise, \( f_y = -4y + 8 \).

Finding these derivatives helps us pinpoint critical points by setting \( f_x \) and \( f_y \) to zero, representing potential points where the function might reach a maximum, minimum, or have a saddle point. In our example, solving \( 2x - 6 = 0 \) and \( -4y + 8 = 0 \) led us to the critical point \( (3, 2) \).

Second Derivative Test

The second derivative test is a method used to classify critical points found using partial derivatives in multivariable calculus. Once we find these critical points, the test helps us determine the nature of these points. Here's how the process works:

For a function \( f(x, y) \), calculate:

• Second partial derivative with respect to \( x \), \( f_{xx} \).
• Second partial derivative with respect to \( y \), \( f_{yy} \).
• Mixed partial derivative \( f_{xy} \).

In our example, these values were \( f_{xx} = 2 \), \( f_{yy} = -4 \), and \( f_{xy} = 0 \).

Then, compute the discriminant \( D \) using the formula: \[ D = f_{xx}f_{yy} - (f_{xy})^2 \].
In our case, \( D = 2(-4) - 0^2 = -8 \).

The sign of \( D \) reveals the point's nature:

• \( D > 0 \): If \( f_{xx} > 0 \), it's a local minimum. If \( f_{xx} < 0 \), it's a local maximum.
• \( D < 0 \): Indicates a saddle point, as seen in our solution where \( D = -8 \).

Saddle Point

A saddle point is a kind of critical point where the function does not have a local maximum or minimum, but instead resembles a saddle or a horse's back. In simpler terms, at a saddle point, the function curves upwards in one direction and downwards in another.

From the second derivative test, we learned that a saddle point occurs when the discriminant \( D < 0 \). In our problem, at the critical point \( (3, 2) \), since \( D = -8 \), it confirms the existence of a saddle point.

Understanding saddle points is important because:

• They represent points where a function flattens out but does not exhibit extremality (neither a peak nor a trough).
• In visual terms, if you were to graph the function, the surface would rise in one direction and fall in another, just like a saddle.

Therefore, although saddle points are flat, they imply that the relationship between the variables around this point is quite dynamic, not stable like a maximum or minimum.