Problem 35 - Calculus Early Transcendentals

Question

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$f(x, y)=2 x^{3}+y^{4}, \quad D=\\{(x, y) | x^{2}+y^{2} \leqslant 1\\}$$

Step-by-Step Solution

Answer

The absolute maximum value is 2, and the absolute minimum value is -2.

Step 1: Understand the problem

We need to find the absolute maximum and minimum values of the function \( f(x, y) = 2x^3 + y^4 \) within the closed set \( D = \{(x, y) \mid x^2 + y^2 \leq 1\} \). The set \( D \) is a closed disk centered at the origin with radius 1.

Step 2: Find critical points inside the region

To find critical points inside \( D \), calculate the partial derivatives: \( f_x = \frac{\partial}{\partial x}(2x^3 + y^4) = 6x^2 \) and \( f_y = \frac{\partial}{\partial y}(2x^3 + y^4) = 4y^3 \).Set these derivatives to zero: \( 6x^2 = 0 \) and \( 4y^3 = 0 \), giving the critical point \( (0, 0) \) within \( D \).

Step 3: Evaluate function at critical points

Substitute the critical point \( (0, 0) \) into \( f(x, y) \): \[ f(0, 0) = 2(0)^3 + (0)^4 = 0 \]

Step 4: Analyze the boundary

On the boundary of \( D \), where \( x^2 + y^2 = 1 \), use a parameterization. Let \( x = \cos(\theta), y = \sin(\theta) \), then \( f(x, y) = 2\cos^3\theta + \sin^4\theta \).Evaluate this expression for \( \theta \) ranging from \( 0 \) to \( 2\pi \).

Step 5: Calculate points on the boundary

Compute \( f(x, y) = 2\cos^3\theta + \sin^4\theta \) by testing values such as \( \theta = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2} \):- At \( \theta = 0 \), \( f(1,0) = 2(1)^3 + (0)^4 = 2 \)- At \( \theta = \frac{\pi}{2} \), \( f(0,1) = 2(0)^3 + (1)^4 = 1 \)- At \( \theta = \pi \), \( f(-1,0) = 2(-1)^3 + (0)^4 = -2 \)- At \( \theta = \frac{3\pi}{2} \), \( f(0,-1) = 2(0)^3 + (-1)^4 = 1 \)

Step 6: Compare values

Compare the values of \( f \) at the critical points and on the boundary: - \( f(0, 0) = 0 \)- Maximum on the boundary: \( f(1,0) = 2 \)- Minimum on the boundary: \( f(-1,0) = -2 \)

Key Concepts

Critical Points

Critical points are essential in calculus optimization problems as they help identify where a function may achieve its maximum or minimum values within a given region. These points occur where the partial derivatives of the function equal zero. For any function of two variables, say, \(f(x, y)\), setting the first derivatives \(f_x\) and \(f_y\) to zero helps find these points.

• Derivatives measure the rate of change of the function with respect to each variable.
• Setting partial derivatives to zero indicates potential stability points.
• The critical points found must lie within the search region—in this example, the region \(D\) defined by the inequality \(x^2 + y^2 \leq 1\).

In the given problem, solving \(6x^2 = 0\) and \(4y^3 = 0\) yields the critical point (0,0). This is a key candidate for checking minimum or maximum values within the set \(D\). Finding these points narrows down the fields to examine further for function analysis.

Partial Derivatives

Partial derivatives are a crucial tool in analyzing functions with more than one variable. They help us understand how the function changes with respect to each independent variable while keeping the other constant. This is integral to finding critical points and performing optimization tasks.

• The partial derivative \(f_x\) is determined by differentiating the function with respect to \(x\), treating \(y\) as a constant.
• Similarly, \(f_y\), is derived by differentiating with respect to \(y\), treating \(x\) as a constant.
• In this exercise: \(f_x = 6x^2\) indicates how \(f\) changes as \(x\) varies, and \(f_y = 4y^3\) shows its change as \(y\) changes.

Calculating these derivatives enables us to analyze and set them to zero for critical point determination. Mastery over partial derivatives is vital as it sets the foundation for deeper optimization analysis.

Boundary Values

In optimization problems, understanding boundary values is indispensable since extrema might occur on the perimeter of the defined space. For regions like circles, squares, or any geometric shape in the coordinate plane, exploring the boundaries is crucial. In this exercise focusing on the region \(D\) described by \(x^2 + y^2 = 1\), it is paramount to evaluate the function on this circle's outer edge.

• Optimization involves checking all potential maximum and minimum points, not just within the interior—boundaries are equally vital.
• When the derivatives inside the region provide inconclusive points, tracking boundary behavior often reveals extrema.

To evaluate the function on the boundary, substitute the parametric representation \(x= \cos(\theta), y=\sin(\theta)\) in \(D\). Direct evaluations at different key angles on the circle help determine where along the edge \(f(x, y)\) may achieve its highest and lowest values, crucial for comprehensive function analysis.

Parameterization

Parameterization is a technique used to express the variables defining a geometric object in terms of a simpler variable, often a single parameter such as \(\theta\), especially useful in problems involving circles and ellipses. This approach simplifies boundary analysis by converting a complex shape into a function of one variable.

• For circles, using polar coordinates offers practical parameterization: \(x = \cos(\theta), y = \sin(\theta)\).
• This reduces the complexity of dealing with an equation in two variables by analyzing behavior solely as a function of \(\theta\).
• In solving the exercise, this methodology allows the boundary \(x^2 + y^2 = 1\) to be examined via function transformations that are more straightforward to analyze over one parameter.

By simplifying the analysis of where extrema might occur, parameterization aids in overlaying the function's expression along the boundary, enabling easier identification of values like maxima and minima. Understanding and applying parameterization is essential in tackling calculus optimization problems effectively.