Problem 336 - Calculus Volume 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. $$f(x, y)=-x^{2}-5 y^{2}+10 x-30 y-62$$

Step-by-Step Solution

Answer

The critical point \((5, -3)\) is a local maximum.

Step 1: Find the first partial derivatives

First, we need to calculate the partial derivatives of the function \( f(x, y) = -x^2 - 5y^2 + 10x - 30y - 62 \). The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = -2x + 10 \) and with respect to \( y \) is \( \frac{\partial f}{\partial y} = -10y - 30 \).

Step 2: Solve for critical points

To find the critical points, set the first partial derivatives equal to zero. Thus:\( -2x + 10 = 0 \) leading to \( x = 5 \).\( -10y - 30 = 0 \) leading to \( y = -3 \).The critical point is at \((5, -3)\).

Step 3: Find the second partial derivatives

Next, calculate the second partial derivatives. \( \frac{\partial^2 f}{\partial x^2} = -2 \), \( \frac{\partial^2 f}{\partial y^2} = -10 \), and \( \frac{\partial^2 f}{\partial x \partial y} = 0 \).

Step 4: Compute the Hessian determinant

The Hessian determinant \( H \) is given by:\[ H = \frac{\partial^2 f}{\partial x^2} \cdot \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \]Substituting the values, we get:\[ H = (-2)(-10) - (0)^2 = 20 \].

Step 5: Determine the nature of the critical point

Using the second derivative test, if \( H > 0 \) and \( \frac{\partial^2 f}{\partial x^2} < 0 \), then the critical point is a local maximum.Here, \( H = 20 > 0 \) and \( \frac{\partial^2 f}{\partial x^2} = -2 < 0 \), which means the point \((5, -3)\) is a local maximum.

Key Concepts

Critical Points

In multivariable calculus, finding critical points is a crucial step in analyzing the behavior of a function of several variables. A critical point occurs where the first partial derivatives of a function are zero. For the function given, we calculated the first partial derivatives to determine these points.

• The derivative with respect to \( x \) results in \(-2x + 10 = 0\).
• Meanwhile, the derivative with respect to \( y \) gives \(-10y - 30 = 0\).

Solving these equations helps us find the values of \( x \) and \( y \) that make the slope of the tangent plane to the graph of the function zero. In this specific problem, setting each partial derivative equal to zero gives us the critical point \((5, -3)\). Understanding critical points helps us describe where the function reaches potential maxima or minima, or perhaps is neither, if it turns out to be a saddle point.

Partial Derivatives

Partial derivatives allow us to examine the rate at which a function changes as we vary one variable while keeping others constant. They are foundational to finding critical points, especially in functions with more than one variable. For our function, \( f(x, y) = -x^2 - 5y^2 + 10x - 30y - 62 \),

• The partial derivative with respect to \( x \) \( \left( \frac{\partial f}{\partial x} = -2x + 10 \right) \) shows how \( f \) changes along the \( x \)-axis.
• Similarly, the partial derivative with respect to \( y \) \( \left( \frac{\partial f}{\partial y} = -10y - 30 \right) \) describes the rate of change along the \( y \)-axis.

These derivatives are each set to zero to identify critical points. Understanding how to find these derivatives can significantly ease the process of analyzing multi-variable functions. They help us break the problem into more manageable parts, focusing on each dimension separately.

Hessian Determinant

The Hessian determinant is a powerful tool in analyzing the nature of critical points for functions of several variables. It involves second partial derivatives and can tell us if a critical point is a local maximum, minimum, or saddle point.To compute the Hessian for our example, we use the second partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} = -2 \)
• \( \frac{\partial^2 f}{\partial y^2} = -10 \)
• \( \frac{\partial^2 f}{\partial x \partial y} = 0 \)

The formula for the Hessian determinant \( H \) in two variables is:\[ H = \left(\frac{\partial^2 f}{\partial x^2}\right) \left(\frac{\partial^2 f}{\partial y^2}\right) - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 \]Substituting the values gives us \( H = 20 \). A positive Hessian determinant, combined with the sign of \( \frac{\partial^2 f}{\partial x^2} \), helps determine whether we have a local maximum or minimum.

Local Maximum

To determine if a critical point is a local maximum, we use both the Hessian determinant and the sign of the second partial derivative with respect to \( x \). Once we find that \( H = 20 > 0 \) and that the second partial derivative with respect to \( x \) is negative:

• \( \frac{\partial^2 f}{\partial x^2} = -2 < 0 \)

We conclude that the critical point \((5, -3)\) is a local maximum. This occurs because the function is concave down in that region. Identifying whether a point is a local maximum is crucial since it represents the highest point in the immediate vicinity. This concept is widely applied in optimization problems where maximum profit or efficiency is sought.