Problem 8 - Applied Calculus

Question

Find all the critical points and determine whether each is a local maximum, local minimum, a saddle point, or none of these. $$f(x, y)=x^{2}+y^{2}+6 x-10 y+8$$

Step-by-Step Solution

Answer

The critical point is \((-3, 5)\), and it is a local minimum.

Step 1: Find the gradient

To locate the critical points, we first need to find the gradient of the function. The gradient of a function \( f(x, y) \) is given by the vector of its partial derivatives. Compute the partial derivatives:\(\frac{\partial f}{\partial x} = 2x + 6\)\(\frac{\partial f}{\partial y} = 2y - 10\)Hence, the gradient is \( abla f(x, y) = \left< 2x + 6, 2y - 10 \right> \).

Step 2: Set the gradient to zero

To find the critical points, set each component of the gradient equal to zero:1. \( 2x + 6 = 0 \)2. \( 2y - 10 = 0 \)Solve these equations:For \( 2x + 6 = 0 \), \( x = -3 \).For \( 2y - 10 = 0 \), \( y = 5 \).Thus, the critical point is \( (x, y) = (-3, 5) \).

Step 3: Compute the Hessian Matrix

To classify the critical point, we need the second partial derivatives to form the Hessian matrix \( H \):\(\frac{\partial^2 f}{\partial x^2} = 2, \quad \frac{\partial^2 f}{\partial x \partial y} = 0\)\(\frac{\partial^2 f}{\partial y^2} = 2, \quad \frac{\partial^2 f}{\partial y \partial x} = 0\)Thus, the Hessian matrix is:\[H = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}\]

Step 4: Evaluate the determinant and classify the critical point

The determinant of the Hessian matrix \( H \) is:\(\det(H) = (2)(2) - (0)(0) = 4\)Since \( \det(H) > 0 \) and \( \frac{\partial^2 f}{\partial x^2} = 2 > 0 \), the condition indicates that the critical point is a local minimum.

Key Concepts

Gradient Vector

The gradient vector is a central concept in multivariable calculus, essential for finding critical points of a function. Essentially, the gradient vector represents the direction of the steepest ascent from any point on the function. For a function of two variables, like \( f(x, y) = x^2 + y^2 + 6x - 10y + 8 \), the gradient is composed of the partial derivatives with respect to each variable.

• The partial derivative with respect to \( x \) is \( \frac{\partial f}{\partial x} = 2x + 6 \).
• The partial derivative with respect to \( y \) is \( \frac{\partial f}{\partial y} = 2y - 10 \).

The gradient vector \( abla f(x, y) \) is then \( \left< 2x + 6, 2y - 10 \right> \). By setting the gradient vector to zero, we find the critical points of the function. This means solving for \( x \) and \( y \) such that each component of the gradient is zero.
Finding where the gradient equals zero helps us locate points where the function may have a maximum, minimum, or saddle point.

Hessian Matrix

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It provides insight into the curvature of a function, which helps in classifying critical points. For the function \( f(x, y) = x^2 + y^2 + 6x - 10y + 8 \), the Hessian matrix is given by:\[H = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2} \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & 2 \end{bmatrix}\]

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

The Hessian helps determine the nature of critical points by evaluating the matrix's determinant and its leading diagonal entries. The determinant is calculated using the formula \( (2)(2) - (0)(0) = 4 \). This particular Hessian structure indicates the critical points are local minima if the determinant is positive and the diagonal elements are positive.

Local Minimum

A local minimum is a point in the domain of a function where the function value is smaller than the values at nearby points. It represents a small "valley" within the function's graph, at least in the immediate vicinity. For the given function in the exercise, after solving the gradient to zero, we get the critical point \( (-3, 5) \).
We classify this point by analyzing the Hessian matrix's determinant:

• Since \( \det(H) = 4 > 0 \), and \( \frac{\partial^2 f}{\partial x^2} = 2 > 0 \), the conditions for a local minimum are satisfied.
• This means that, around \( (-3, 5) \), the function \( f(x, y) \) decreases in any direction, indicating a local minimum.

In simple terms, the graph of the function dips at \( (-3, 5) \), forming a trough or small valley, where it achieves its lowest value in that region.

Partial Derivatives

Partial derivatives are foundational components in calculus used to analyze functions of multiple variables. They measure how a function changes as one variable changes while keeping other variables constant. For instance, in the function provided \( f(x, y) = x^2 + y^2 + 6x - 10y + 8 \), we calculate partial derivatives:

• \( \frac{\partial f}{\partial x} = 2x + 6 \)
• \( \frac{\partial f}{\partial y} = 2y - 10 \)

Each derivative only considers the change with respect to its variable, offering a snapshot of the function’s slope along the axes.
Using partial derivatives, we derive the gradient vector, which helps find critical points where the function’s slope is zero, indicating potential extremums. Mastering the concept of partial derivatives is crucial for studying the behavior and characteristics of complex functions involving multiple variables.