Problem 326 - Calculus 3

Question

For the following exercises, use the second derivative test to identify any critical points and determine whether each critical point is a maximum, minimum, saddle point, or none of these. $$ f(x, y)=7 x^{2} y+9 x y^{2} $$

Step-by-Step Solution

Answer

The critical point is \((0, 0)\) and its nature is inconclusive by the second derivative test.

Step 1: Calculate First Partial Derivatives

To find the critical points, begin by calculating the first partial derivatives of the function \( f(x, y) \). The partial derivative with respect to \( x \) is \( f_x = \frac{\partial}{\partial x}(7x^2y + 9xy^2) = 14xy + 9y^2 \). The partial derivative with respect to \( y \) is \( f_y = \frac{\partial}{\partial y}(7x^2y + 9xy^2) = 7x^2 + 18xy \).

Step 2: Find Critical Points

To find critical points, set both partial derivatives equal to zero and solve for \( x \) and \( y \). For \( f_x = 0 \), we have \( 14xy + 9y^2 = 0 \), which can be factored as \( y(14x + 9y) = 0 \). For \( f_y = 0 \), solve \( 7x^2 + 18xy = 0 \), which factors to \( x(7x + 18y) = 0 \). Solving these equations simultaneously gives critical points \((0, 0)\) and \((0, 0)\).

Step 3: Calculate Second Partial Derivatives

Next, calculate the second partial derivatives: \( f_{xx} = \frac{\partial^2}{\partial x^2} (7x^2y + 9xy^2) = 14y \), \( f_{yy} = \frac{\partial^2}{\partial y^2} (7x^2y + 9xy^2) = 18x \), and \( f_{xy} = \frac{\partial^2}{\partial x \partial y} (7x^2y + 9xy^2) = 14x + 18y \).

Step 4: Apply Second Derivative Test

For the second derivative test, compute the Hessian determinant \( D = f_{xx}f_{yy} - (f_{xy})^2 \) at each critical point. For the critical point \((0, 0)\), compute \( f_{xx} = 14 \cdot 0 = 0 \), \( f_{yy} = 18 \cdot 0 = 0 \), and \( f_{xy} = 0 \). Thus, the Hessian determinant \( D = 0 \cdot 0 - 0^2 = 0 \). The second derivative test is inconclusive when \( D = 0 \), so the nature of this critical point cannot be determined from this test.

Key Concepts

Critical Points

Critical points in calculus are specific points on a function's graph where something interesting happens. Generally, these points occur where the derivative(s) of the function are zero or undefined. For functions of two variables, like our example, a critical point \( (x, y) \) is a place where the first partial derivatives \( f_{x}(x, y) \) and \( f_{y}(x, y) \) are both equal to zero.

In simple terms, critical points could be where the function has a peak (maximum), a valley (minimum), or a saddle point. To find these, you'll first calculate the partial derivatives then set each of them to zero. This helps locate points where the function might change direction.

For instance, in our exercise, we found the critical point \( (0, 0) \) by setting the partial derivatives \( 14xy + 9y^{2} = 0 \) and \( 7x^{2} + 18xy = 0 \) to zero. Solving these simultaneously reveals the specific values of \( x \) and \( y \) at which the function behaves differently.

Partial Derivatives

Partial derivatives serve as a crucial tool, especially when handling functions with multiple variables. These derivatives help describe how the function changes as you tweak one variable while keeping the others constant.

Imagine having a function \( f(x, y) \). The partial derivative with respect to \( x \) (denoted as \( f_x \)) can be thought of as "how quickly does \( f \) change if \( x \) changes just a little, while \( y \) stays put?" Similarly, \( f_y \) is the change in \( f \) as \( y \) changes independently.

In our function \( f(x, y)=7x^2y+9xy^2 \), we calculated \( f_x = 14xy + 9y^2 \) and \( f_y = 7x^2 + 18xy \). These equations indicate how the function's slope changes based on altering \( x \) or \( y \) and eventually help us locate the critical points by setting them to zero.

Hessian Determinant

The Hessian determinant is a vital component used in the second derivative test to understand the nature of critical points for functions with multiple variables. This determinant involves the second partial derivatives of a function. It's best described as a single value derived from these derivatives, offering insights on whether a critical point is a maximum, minimum, or a saddle point.

To compute the Hessian determinant \( D \) for a function \( f(x, y) \), use the formula: \[D = f_{xx} f_{yy} - (f_{xy})^2\]where:

• \( f_{xx} \) is the second partial derivative with respect to \( x \)
• \( f_{yy} \) is the second partial derivative with respect to \( y \)
• \( f_{xy} \) is the mixed derivative, reflecting changes when both \( x \) and \( y \) modify.

In the exercise, we calculated \( D = 0 \) at the critical point \( (0,0) \). Since \( D = 0 \), the second derivative test does not provide any definitive conclusion about the nature of this critical point.