Problem 33 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$\begin{array}{l}{f(x, y)=x^{4}+y^{4}-4 x y+2} \\ {D=\\{(x, y) | 0 \leqslant x \leqslant 3,0 \leqslant y \leqslant 2\\}}\end{array}$$

Step-by-Step Solution

Answer

The absolute maximum is 83 at (3, 0); the absolute minimum is 0 at (1, 1).

Step 1: Find Critical Points

To find the critical points of the function \(f(x, y) = x^4 + y^4 - 4xy + 2\), we first compute the partial derivatives with respect to \(x\) and \(y\). The partial derivatives are: \(f_x(x, y) = 4x^3 - 4y\) and \(f_y(x, y) = 4y^3 - 4x\). Set each partial derivative to zero to find the critical points: \(4x^3 - 4y = 0\) and \(4y^3 - 4x = 0\). Solving these equations, we get \(x^3 = y\) and \(y^3 = x\). Solving for \(x\) in terms of \(y\), we have \(x = y^{1/3}\) and \(y = x^{1/3}\). The only real solution to both is \(x = y = 1\), which indicates one critical point at \((1, 1)\).

Step 2: Evaluate Function at Critical Points

Evaluate \(f\) at the critical point \((1, 1)\): \[f(1, 1) = 1^4 + 1^4 - 4(1)(1) + 2 = 2 - 4 + 2 = 0.\] So, \(f(1, 1) = 0\).

Step 3: Evaluate Function on Boundary

Now check function values at the boundaries of \(D\):1. For \(x = 0\), evaluate \(f(0, y) = y^4 + 2\) at the endpoints \(y = 0\) and \(y = 2\): - \(f(0, 0) = 0^4 + 2 = 2\) - \(f(0, 2) = 2^4 + 2 = 18\) 2. For \(x = 3\), evaluate \(f(3, y) = 3^4 + y^4 - 12y + 2\) at \(y = 0\) and \(y = 2\): - \(f(3, 0) = 3^4 + 0^4 - 12(0) + 2 = 81 + 2 = 83\) - \(f(3, 2) = 3^4 + 2^4 - 12(2) + 2 = 81 + 16 - 24 + 2 = 75\)3. For \(y = 0\), evaluate \(f(x, 0) = x^4 + 2\) at \(x = 0\) and \(x = 3\): - \(f(0, 0) = 0^4 +2 = 2\) - \(f(3, 0) = 3^4 + 2 = 83\)4. For \(y = 2\), evaluate \(f(x, 2) = x^4 + 16 - 8x + 2\) at \(x = 0\) and \(x = 3\): - \(f(0, 2) = 0^4 + 16 - 8(0) + 2 = 18\) - \(f(3, 2) = 3^4 + 16 - 8(3) + 2 = 75\)

Step 4: Compare Values

Compare all calculated values to determine the absolute maximum and minimum:- Critical Point: \(f(1, 1) = 0\)- Boundary Points: - \(f(0, 0) = 2\), \(f(0, 2) = 18\), \(f(3, 0) = 83\), \(f(3, 2) = 75\)The smallest value is \(f(1, 1) = 0\) and the largest value is \(f(3, 0) = 83\).

Key Concepts

Critical Points

Critical points are essential for finding absolute extrema, which are the highest or lowest values a function can have in a given domain. To find the critical points of a multivariable function like the one given, we first compute the partial derivatives of the function. For the function \(f(x, y) = x^4 + y^4 - 4xy + 2\), the partial derivatives with respect to\(x\) and\(y\) are:

• \( f_x(x, y) = 4x^3 - 4y \)
• \( f_y(x, y) = 4y^3 - 4x \)

A critical point occurs where these partial derivatives are zero. By setting \( 4x^3 - 4y = 0 \) and \( 4y^3 - 4x = 0 \), we solve for \(x\) and \(y\) to find the critical points. Solving the system of equations leads to \(x = y = 1\), indicating a critical point at \((1, 1)\). This critical point will be evaluated later to contribute towards finding absolute extrema.

Boundary Evaluation

Boundary evaluation involves analyzing the function values at the edges or limits of the given domain \(D\). For the exercise, the domain is \(D= \{ (x, y) | 0 \leq x \leq 3, 0 \leq y \leq 2\} \). Check the function where \(x\) and \(y\) take boundary values:

• When \(x = 0\), then evaluate \(f(0, y) = y^4 + 2\) at \(y = 0\) and \(y = 2\).
• When \(x = 3\), evaluate \(f(3, y) = 3^4 + y^4 - 12y + 2\) at \(y = 0\) and \(y = 2\).
• When \(y = 0\), evaluate \(f(x, 0) = x^4 + 2\) at \(x = 0\) and \(x = 3\).
• When \(y = 2\), evaluate \(f(x, 2) = x^4 + 16 - 8x + 2\) at \(x = 0\) and \(x = 3\).

Each evaluation provides potential extrema candidates on the boundary. These need to be compared to those obtained from the critical points.

Partial Derivatives

Partial derivatives are crucial in the investigation of multivariable functions, allowing us to understand how the function changes with respect to each variable independently. For the function \(f(x, y) = x^4 + y^4 - 4xy + 2\), partial derivatives were found as:

• \(f_x(x, y) = \frac{\partial f}{\partial x} = 4x^3 - 4y\)
• \(f_y(x, y) = \frac{\partial f}{\partial y} = 4y^3 - 4x\)

These derivatives are set to zero to locate critical points within the domain. Solving these equations respects the interaction between \(x\) and \(y\), providing insights into behaviors like stationary points, which are potential locations for maximum or minimum values.

Function Evaluation

Function evaluation is the process of determining the function's output values at specific points, particularly at critical points and boundaries identified previously. In this example, we evaluate the function:

• At the critical point \((1, 1)\), the function value is \(f(1, 1) = 0\).
• At the boundary points, function values range; for instance, \(f(0, 2) = 18\), \(f(3, 0) = 83\), among others.

These evaluations show which point corresponds to the highest and lowest values within the domain. The maximum value identified is at \((3, 0)\) with \(f(3, 0) = 83\), and the minimum is at the critical point \((1, 1)\) where \(f(1, 1) = 0\). This thorough examination confirms the absolute extrema on the set \(D\).