Problem 12 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the global maximum value and global minimum value of \(f\) on \(S\) and indicate where each occurs. \(f(x, y)=x^{2}+y^{2}\); \(S=\\{(x, y):-1 \leq x \leq 3,-1 \leq y \leq 4\\}\)

Step-by-Step Solution

Answer

Global maximum is 25 at (3,4); global minimum is 0 at (0,0).

Step 1: Understand the Function and Domain

We have the function \(f(x, y) = x^2 + y^2\) and the domain \(S = \{(x, y) : -1 \leq x \leq 3, -1 \leq y \leq 4\}\). This domain describes a rectangle in the xy-plane.

Step 2: Identify Critical Points Inside the Domain

To find critical points, compute the partial derivatives \(f_x = 2x\) and \(f_y = 2y\). Set them equal to zero, giving \(x = 0\) and \(y = 0\). The critical point is \((0, 0)\). Check if it lies within the domain: \(-1 \leq 0 \leq 3\) and \(-1 \leq 0 \leq 4\); hence \((0, 0)\) is inside \(S\).

Step 3: Evaluate the Function at the Critical Point

At the critical point \((0, 0)\), \(f(0, 0) = 0^2 + 0^2 = 0\).

Step 4: Evaluate the Function Along the Boundaries of the Domain

Examine the boundaries: \(x = -1\), \(x = 3\), \(y = -1\), and \(y = 4\). Calculate \(f\) at corners: - \((-1, -1): f(-1, -1) = (-1)^2 + (-1)^2 = 2\)- \((-1, 4): f(-1, 4) = (-1)^2 + 4^2 = 17\)- \((3, -1): f(3, -1) = 3^2 + (-1)^2 = 10\)- \((3, 4): f(3, 4) = 3^2 + 4^2 = 25\)

Step 5: Compare Values to Find Global Maximum and Minimum

Compare \(f(0, 0) = 0\), \(f(-1, -1) = 2\), \(f(-1, 4) = 17\), \(f(3, -1) = 10\), and \(f(3, 4) = 25\). The global maximum is \(25\) at point \((3, 4)\) and the global minimum is \(0\) at \((0, 0)\).

Key Concepts

Critical Points

In calculus, "critical points" of a function are where the function's derivative is zero or undefined. These points help to determine potential maxima, minima, or saddle points. For a function of two variables like in our exercise, we find critical points using partial derivatives.

To locate critical points:

• Compute the partial derivatives of the function with respect to each variable.
• Set these derivatives equal to zero.
• Solve the equations simultaneously to find critical points.

In the example problem, the function is given by \(f(x, y) = x^2 + y^2\). The partial derivatives \(f_x = 2x\) and \(f_y = 2y\) are set to zero to find values which make the derivative zero.

This leads to the critical point \((0, 0)\), which lies within the given domain. Once found, it’s crucial to evaluate the function at these points to check if they are indeed global extremum points.

Partial Derivatives

Partial derivatives are crucial in finding critical points of multivariable functions. They measure how a function changes as each variable changes, treating all other variables as constants.

Why are partial derivatives important?

• They help in understanding the behavior of the function with respect to each variable independently.
• In optimization problems, they are used to find points where the function's slope is zero, indicating potential maxima, minima, or saddle points.

For our function \(f(x, y) = x^2 + y^2\), the partial derivative with respect to \(x\) is \(f_x = 2x\) and with respect to \(y\) is \(f_y = 2y\). Setting these equal to zero helps identify the critical point as \((0, 0)\). These derivatives tell us how \(f\) changes with changes in \(x\) and \(y\).

Boundary Evaluation

Boundary evaluation involves analyzing the function on the perimeter or limits of the domain. This step is crucial when looking for global extrema, as potential maxima or minima can occur at boundaries.

In optimization, boundary evaluation ensures that all possible extremum points are considered, not just those within the interior of the domain.

• Identify the corners or edges of the domain and assess the function at these points.
• Compare these boundary values with those found at internal critical points.

For the function at hand, the domain is a rectangle, bounded by lines \(x = -1\), \(x = 3\), \(y = -1\), and \(y = 4\). Evaluating \(f\) at the corners — \((-1, -1), (-1, 4), (3, -1), (3, 4)\) — provides values that must be compared to those at the critical points.

This process reveals the global maximum of 25 at \((3, 4)\) and global minimum of 0 at \((0, 0)\), thus ensuring a comprehensive search for extrema.