14.7: Global Optimization Find the absolute maximum and minimum values of the function f (x, y ) = xy 2 on the domain D = {(x, y )|x ≥ 0, y ≤ 0, x2 + y 2 ≤ 9} can you use a table or 2nd derivative test table if you can

To find the absolute maximum and minimum values of \( f(x, y) = xy^2 \) over the domain \( D = \{(x, y) | x \geq 0, y \leq 0, x^2 + y^2 \leq 9\} \), we’ll examine critical points within the interior of \( D \) and analyze the behavior of \( f \) on the boundary.

### Step 1: Define and Analyze Critical Points

#### 1. Compute the partial derivatives of \( f \):
\[
f_x = \frac{\partial}{\partial x} (xy^2) = y^2
\]
\[
f_y = \frac{\partial}{\partial y} (xy^2) = 2xy
\]

#### 2. Set the partial derivatives to zero to find the critical points:
For \( f_x = 0 \), we have \( y^2 = 0 \Rightarrow y = 0 \).
For \( f_y = 0 \), we have \( 2xy = 0 \).

Since \( y = 0 \), the only solution is \( x = 0 \). Thus, the only critical point in the interior of \( D \) is at \( (0, 0) \).

### Step 2: Evaluate \( f \) at the Critical Point
\[
f(0, 0) = 0
\]

### Step 3: Examine the Boundary of \( D \)

The boundary of \( D \) is defined by \( x^2 + y^2 = 9 \) with \( x \geq 0 \) and \( y \leq 0 \). Parameterize this boundary using polar coordinates where \( x = 3 \cos \theta \) and \( y = 3 \sin \theta \), with \( \pi \leq \theta \leq \frac{3\pi}{2} \).

Substitute these into \( f(x, y) = xy^2 \):
\[
f(3 \cos \theta, 3 \sin \theta) = (3 \cos \theta)(3 \sin \theta)^2 = 3 \cos \theta \cdot 9 \sin^2 \theta = 27 \cos \theta \sin^2 \theta
\]

To find the maximum and minimum, we need to analyze \( 27 \cos \theta \sin^2 \theta \) on the interval \( \pi \leq \theta \leq \frac{3\pi}{2} \).

#### Rewrite Using Trigonometric Identity
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \):
\[
f(3 \cos \theta, 3 \sin \theta) = 27 \cos \theta (1 - \cos^2 \theta)
\]

Define \( u = \cos \theta \), where \( -1 \leq u \leq 0 \). Then \( f(u) = 27 u (1 - u^2) = 27(u - u^3) \).

#### Differentiate \( f(u) = 27(u - u^3) \):
\[
f'(u) = 27(1 - 3u^2)
\]
Set \( f'(u) = 0 \):
\[
1 - 3u^2 = 0 \Rightarrow u^2 = \frac{1}{3} \Rightarrow u = -\frac{1}{\sqrt{3}}
\]

#### Evaluate \( f(u) \) at Critical Points and Endpoints
1. **At \( u = -\frac{1}{\sqrt{3}} \):**
   \[
   f\left(-\frac{1}{\sqrt{3}}\right) = 27\left(-\frac{1}{\sqrt{3}} - \left(-\frac{1}{\sqrt{3}}\right)^3\right) = 27\left(-\frac{1}{\sqrt{3}} + \frac{1}{3\sqrt{3}}\right) = -18
   \]

2. **At \( u = -1 \):**
   \[
   f(-1) = 27(-1 - (-1)^3) = 0
   \]

3. **At \( u = 0 \):**
   \[
   f(0) = 0
   \]

### Conclusion
The absolute maximum value of \( f(x, y) \) on \( D \) is \( 0 \) and the absolute minimum value is \( -18 \). These occur at:
- **Maximum** at \( (0, 0) \) and on the boundary where \( u = 0 \) or \( u = -1 \).
- **Minimum** at \( u = -\frac{1}{\sqrt{3}} \).

Would you like more detailed steps on solving this or to explore similar problems?

### Follow-Up Questions:
1. How does the boundary of a domain affect finding the extrema of a function?
2. What role does parameterization play in boundary evaluations for optimization?
3. How can polar coordinates simplify the analysis of circular regions?
4. When is the second derivative test useful in finding absolute extrema?
5. How do critical points in a function’s interior compare to those on the boundary?

### Tip:
When analyzing extrema on a domain, remember to always consider both interior points and boundary points.

14.7: Global Optimization Find the absolute maximum and minimum values of the function f(x, y) = xy^2 on the domain D = {(x, y) | x ≥ 0, y ≤ 0, x^2 + y^2 ≤ 9}. Can you use a table or 2nd derivative test table if you can.

Discover the optimization of the function f(x, y) = xy^2 over a bounded domain D defined by x ≥ 0, y ≤ 0, and x^2 + y^2 ≤ 9. This solution involves critical point analysis, boundary parameterization using polar coordinates, and detailed calculations to find absolute maximum and minimum values. Suitable for multivariable calculus students.

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