Find The Absolute Maximum And Minimum Values Of The Function F ( X , Y ) = X Y 2 − Y 2 − X F(x, Y) = Xy^{2}-y^{2}-x F ( X , Y ) = X Y 2 − Y 2 − X On The Region D D D

Jun 16, 2025 by ADMIN 166 views

Finding Absolute Maximum and Minimum Values of Functions on a Region

Finding the absolute maximum and minimum values of a function over a given region is a fundamental problem in multivariable calculus. This article will explore how to solve such problems, focusing on the function $f(x, y) = xy^2 - y^2 - x$ on the region $D = {(x, y) \mid x \geq 0, y \geq 0, x^2 + y^2 \leq 9}$ . This region $D$ is the quarter-disk in the first quadrant with a radius of 3. We will cover the process of finding critical points, analyzing the boundary, and determining the absolute extrema.

Understanding the Problem

Before diving into the solution, it's crucial to understand the problem conceptually. We are given a function $f(x, y)$ and a region $D$ in the $xy$ -plane. Our goal is to find the points in $D$ where $f(x, y)$ attains its highest (absolute maximum) and lowest (absolute minimum) values. These points could be inside the region (critical points) or on the boundary of the region. The Extreme Value Theorem guarantees that a continuous function on a closed and bounded region attains both an absolute maximum and an absolute minimum.

Step 1: Finding Critical Points Inside the Region

To find the critical points inside the region $D$ , we need to find the points where the gradient of $f$ is zero or undefined. The gradient of $f$ is given by the vector of its partial derivatives:

$\nabla f(x, y) = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)$

For our function $f(x, y) = xy^2 - y^2 - x$ , we compute the partial derivatives:

$\frac{\partial f}{\partial x} = y^2 - 1$
$\frac{\partial f}{\partial y} = 2xy - 2y = 2y(x - 1)$

To find the critical points, we set both partial derivatives equal to zero:

$y^2 - 1 = 0$ => $y = \pm 1$
$2y(x - 1) = 0$ => $y = 0$ or $x = 1$

Since we are considering the region where $y \geq 0$ , we only take $y = 1$ and $y = 0$ . Now, we consider the cases:

If $y = 1$ , then from the second equation, $x = 1$ . So, one critical point is $(1, 1)$ .
If $y = 0$ , then the first equation gives $0^2 - 1 = -1 \neq 0$ , which is a contradiction. So, there are no critical points when $y = 0$ .
If $x = 1$ , the first equation gives $y = \pm 1$ , and again we take $y = 1$ since $y \geq 0$ , yielding the critical point $(1, 1)$ .

The critical point $(1, 1)$ lies within our region $D$ because $1 \geq 0$ , $1 \geq 0$ , and $1^2 + 1^2 = 2 \leq 9$ . We evaluate the function at this point:

$f(1, 1) = (1)(1)^2 - (1)^2 - 1 = 1 - 1 - 1 = -1$

Therefore, the critical point inside the region is (1, 1), and the function value at this point is -1. It is important to highlight the calculations in this step as this is a crucial stage in finding the absolute extrema. The partial derivatives must be computed accurately, and the solutions to the system of equations should be verified within the domain.

Step 2: Analyzing the Boundary of the Region

The boundary of the region $D$ consists of three parts:

The line segment along the $x$ -axis from $(0, 0)$ to $(3, 0)$ .
The line segment along the $y$ -axis from $(0, 0)$ to $(0, 3)$ .
The quarter-circle $x^2 + y^2 = 9$ in the first quadrant.

We will analyze each part separately.

2.1: Along the x-axis (y = 0, 0 ≤ x ≤ 3)

On this segment, $y = 0$ , so the function becomes:

$f(x, 0) = x(0)^2 - (0)^2 - x = -x$

This is a simple linear function. On the interval $0 \leq x \leq 3$ , the function decreases as $x$ increases. Thus, the maximum value is at $x = 0$ , and the minimum value is at $x = 3$ .

$f(0, 0) = -0 = 0$
$f(3, 0) = -3$

2.2: Along the y-axis (x = 0, 0 ≤ y ≤ 3)

On this segment, $x = 0$ , so the function becomes:

$f(0, y) = (0)y^2 - y^2 - 0 = -y^2$

This is a quadratic function. On the interval $0 \leq y \leq 3$ , the function decreases as $|y|$ increases. Thus, the maximum value is at $y = 0$ , and the minimum value is at $y = 3$ .

$f(0, 0) = -0^2 = 0$
$f(0, 3) = -3^2 = -9$

2.3: Along the Quarter-Circle (x² + y² = 9, x ≥ 0, y ≥ 0)

To analyze the function on the quarter-circle, we can use the parameterization $x = 3\cos(\theta)$ and $y = 3\sin(\theta)$ , where $0 \leq \theta \leq \frac{\pi}{2}$ . Substituting these into $f(x, y)$ , we get:

$g(\theta) = f(3\cos(\theta), 3\sin(\theta)) = (3\cos(\theta))(3\sin(\theta))^2 - (3\sin(\theta))^2 - 3\cos(\theta)$

$g(\theta) = 27\cos(\theta)\sin^2(\theta) - 9\sin^2(\theta) - 3\cos(\theta)$

Now, we find the critical points of $g(\theta)$ by taking its derivative with respect to $\theta$ and setting it to zero:

$g'(\theta) = -27\sin^3(\theta) + 54\cos^2(\theta)\sin(\theta) - 18\sin(\theta)\cos(\theta) + 3\sin(\theta)$

$g'(\theta) = 3\sin(\theta)(-9\sin^2(\theta) + 18\cos^2(\theta) - 6\cos(\theta) + 1)$

Setting $g'(\theta) = 0$ , we have two cases:

$\sin(\theta) = 0$ => $\theta = 0$
$-9\sin^2(\theta) + 18\cos^2(\theta) - 6\cos(\theta) + 1 = 0$

For the second case, we use the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ to rewrite the equation in terms of $\cos(\theta)$ :

$-9(1 - \cos^2(\theta)) + 18\cos^2(\theta) - 6\cos(\theta) + 1 = 0$

$27\cos^2(\theta) - 6\cos(\theta) - 8 = 0$

Let $u = \cos(\theta)$ . Then we have a quadratic equation:

$27u^2 - 6u - 8 = 0$

Using the quadratic formula:

$u = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(27)(-8)}}{2(27)} = \frac{6 \pm \sqrt{36 + 864}}{54} = \frac{6 \pm \sqrt{900}}{54} = \frac{6 \pm 30}{54}$

So, $u = \frac{36}{54} = \frac{2}{3}$ or $u = \frac{-24}{54} = -\frac{4}{9}$ .

Since $0 \leq \theta \leq \frac{\pi}{2}$ , we have $0 \leq \cos(\theta) \leq 1$ , so we consider $u = \frac{2}{3}$ .

$\cos(\theta) = \frac{2}{3}$ => $\theta = \arccos(\frac{2}{3})$

If $\cos(\theta) = \frac{2}{3}$ , then $\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$ .

Thus, $x = 3\cos(\theta) = 3(\frac{2}{3}) = 2$ and $y = 3\sin(\theta) = 3(\frac{\sqrt{5}}{3}) = \sqrt{5}$ .

Now, we evaluate $f(x, y)$ at the critical points on the boundary:

At $(3,0)$ : $f(3, 0) = -3$
At $(0,3)$ : $f(0, 3) = -9$
At $(2, \sqrt{5})$ : $f(2, \sqrt{5}) = 2(\sqrt{5})^2 - (\sqrt{5})^2 - 2 = 2(5) - 5 - 2 = 10 - 5 - 2 = 3$

Analyzing the boundary is the most complex part, particularly the quarter-circle. The use of trigonometric parameterization and the computation of the derivative of the resulting function $g( heta)$ are essential. Solving for $\theta$ and converting back to $x$ and $y$ requires careful attention to detail. The boundary analysis ensures that we cover all potential extrema along the edges of the region.

Step 3: Comparing Values and Finding Absolute Extrema

We have the following function values at critical points and boundary points:

$f(1, 1) = -1$ (critical point inside $D$ )
$f(0, 0) = 0$ (boundary)
$f(3, 0) = -3$ (boundary)
$f(0, 3) = -9$ (boundary)
$f(2, \sqrt{5}) = 3$ (boundary)

Comparing these values, we find:

Absolute Maximum: $f(2, \sqrt{5}) = 3$
Absolute Minimum: $f(0, 3) = -9$

Therefore, the absolute maximum value of the function $f(x, y) = xy^2 - y^2 - x$ on the region $D$ is 3, which occurs at the point $(2, \sqrt{5})$ , and the absolute minimum value is -9, which occurs at the point $(0, 3)$ . It's crucial to emphasize that comparing all critical points, including those on the boundary, is the final step in identifying the absolute extrema.

Conclusion

Finding the absolute maximum and minimum values of a function on a region involves several steps: finding critical points inside the region, analyzing the boundary, and comparing the function values at all critical points and endpoints. For the function $f(x, y) = xy^2 - y^2 - x$ on the region $D = \{(x, y) \mid x \geq 0, y \geq 0, x^2 + y^2 \leq 9\}$ , we found that the absolute maximum value is 3, occurring at $(2, \sqrt{5})$ , and the absolute minimum value is -9, occurring at $(0, 3)$ .

This comprehensive approach ensures that all potential extrema are considered, leading to the correct identification of the absolute maximum and minimum values. Mastering these techniques is essential for solving a wide range of optimization problems in various fields of science and engineering.