Problem 346 - Calculus Volume 3

Question

Find the absolute extrema of the given function on the indicated closed and bounded set \(R\). \(\quad f(x, y)=x^{3}-3 x y-y^{3}\) on \(R=\\{(x, y):-2 \leq x \leq 2,-2 \leq y \leq 2\\}\).

Step-by-Step Solution

Answer

The absolute minimum is \(-24\) and the absolute maximum is \(24\).

Step 1: Understand the Problem

We need to find the absolute maximum and minimum values of the function \( f(x, y) = x^3 - 3xy - y^3 \) on the region \( R = \{ (x, y) : -2 \leq x \leq 2, -2 \leq y \leq 2 \} \), which is a square in the \(xy\)-plane.

Step 2: Find Critical Points Inside the Region

Calculate the partial derivatives: \( f_x = 3x^2 - 3y \) and \( f_y = -3x - 3y^2 \). Solve \( f_x = 0 \) and \( f_y = 0 \) simultaneously.Solving these equations:- \( f_x = 3x^2 - 3y = 0 \) \( \Rightarrow y = x^2 \)- \( f_y = -3x - 3y^2 = 0 \) \( \Rightarrow x = -y^2 \)Substitute \( y = x^2 \) into \( x = -y^2 \):- \( x = -x^4 \)- Leads to \( x(x^3 + 1) = 0 \)- Solutions are \( x = 0 \) or no real solutions for \( x^3 + 1 = 0 \)Check \( x = 0 \):- \( y = x^2 = 0 \)- Critical point: \( (0, 0) \).

Step 3: Evaluate the Function at the Critical Point

Evaluate \( f(0, 0) = 0^3 - 3(0)(0) - 0^3 = 0 \).

Step 4: Evaluate the Function on the Boundary of the Region

Evaluate \( f \) on the boundary of region \( R \), which includes checking all edge cases:1. **Edge:** \( x = -2 \) then \(-2 \leq y \leq 2\), \( f(-2, y) = -8 + 6y - y^3 \) - Compute endpoints: \( f(-2, -2) = -8 \), \( f(-2, 2) = 24 \).2. **Edge:** \( x = 2 \) then \(-2 \leq y \leq 2\), \( f(2, y) = 8 - 6y - y^3 \) - Compute endpoints: \( f(2, -2) = 8 \), \( f(2, 2) = -24 \).3. **Edge:** \( y = -2 \) then \(-2 \leq x \leq 2\), \( f(x, -2) = x^3 + 6x + 8 \) - Compute endpoints: \( f(-2, -2) = -8 \), \( f(2, -2) = 24 \).4. **Edge:** \( y = 2 \) then \(-2 \leq x \leq 2\), \( f(x, 2) = x^3 - 6x - 8 \) - Compute endpoints: \( f(-2, 2) = 8 \), \( f(2, 2) = -24 \).

Step 5: Compare the Values

Compare all the function values calculated:- Critical point: \( f(0, 0) = 0 \).- Boundary points: \( -24, 0, 8, 24 \).Find the minimum and maximum values:- Absolute minimum: \( -24 \).- Absolute maximum: \( 24 \).

Key Concepts

Partial Derivatives

Partial derivatives are crucial in finding extrema for functions of multiple variables. In our scenario, we have a function of two variables, \( f(x, y) = x^3 - 3xy - y^3 \).
To determine how \( f \) changes with respect to one variable while keeping the other constant, we calculate partial derivatives. These derivatives are like regular derivatives but focus on one direction in the multi-dimensional space at a time.

For our function:

• The partial derivative with respect to \( x \) is \( f_x = \frac{\partial f}{\partial x} = 3x^2 - 3y \).
• The partial derivative with respect to \( y \) is \( f_y = \frac{\partial f}{\partial y} = -3x - 3y^2 \).

These help us to find where the function's rate of change is zero in each direction, leading us to critical points inside the bounded region.

Critical Points

Critical points are where the function potentially achieves its highest or lowest values within a region.
They are found by setting the partial derivatives to zero, which means the slope of the function is level at those points, indicating a possibility of an extremum.

For the function \( f(x, y) \), we solve the system:

• \( f_x = 0 \Rightarrow 3x^2 - 3y = 0 \Rightarrow y = x^2 \)
• \( f_y = 0 \Rightarrow -3x - 3y^2 = 0 \Rightarrow x = -y^2 \)

By substituting \( y = x^2 \) into \( x = -y^2 \), we solve \( x = -x^4 \), getting \( x = 0 \). Then \( y = x^2 = 0 \) gives us the critical point \( (0, 0) \). This is the only critical point within our region of interest, \((-2 \leq x, y \leq 2)\).

Bounded Region

When identifying absolute extrema, consideration must be given to the behavior of the function over the entire region, including its boundaries.
In our problem, the region \( R \) is a closed square defined by the constraints \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\). This means our function is evaluated not just at the interior critical points, but also along the edges of this square.

On each edge:

• For \( x = -2 \) or \( x = 2 \), we evaluate \( f \) as a function of \( y \).
• For \( y = -2 \) or \( y = 2 \), we evaluate \( f \) as a function of \( x \).

These evaluations ensure no potential extrema along the boundaries are overlooked, confirming any absolute extrema within the bounded region.

Function Evaluation

Evaluating the function appropriately at all key points helps in identifying the absolute extrema.
We need to compare the function's value at the critical points and along the boundary's edges to determine where it is maximized or minimized.

Steps in the evaluation:

• At the critical point \( (0, 0) \), \( f(0, 0) = 0 \).
• On the boundary
  • For \( x = -2 \): calculate at \( y = -2 \) and \( y = 2 \).
  • For \( x = 2 \): check \( y = -2 \) and \( y = 2 \).
  • For \( y = -2 \): evaluate at \( x = -2 \) and \( x = 2 \).
  • For \( y = 2 \): check \( x = -2 \) and \( x = 2 \).

Through these calculations, we find function values at boundary points and critical points, ensuring we don't miss the absolute maximum (24) or minimum (-24) values within the specified region.