Find the absolute maximum and minimum values of *f* on the set *D*.

*f*(*x*, *y*) = *xy*2 + 9,    *D* = {(*x*, *y*) | *x* ≥ 0, *y* ≥ 0, *x*2 + *y*2 ≤ 3} Give your answer in terms of exact numbers

We are tasked with finding the absolute maximum and minimum values of the function \( f(x, y) = xy^2 + 9 \) on the set \( D = \{(x, y) \ | \ x \geq 0, y \geq 0, x^2 + y^2 \leq 3\} \), which is a quarter-circle in the first quadrant of the plane.

### Step 1: Critical points in the interior of the domain
To find critical points in the interior of \( D \), we first compute the partial derivatives of \( f(x, y) \):

\[
f_x(x, y) = \frac{\partial}{\partial x}(xy^2 + 9) = y^2
\]
\[
f_y(x, y) = \frac{\partial}{\partial y}(xy^2 + 9) = 2xy
\]

To find critical points, we set these partial derivatives equal to zero:
1. \( f_x(x, y) = y^2 = 0 \)
2. \( f_y(x, y) = 2xy = 0 \)

From \( y^2 = 0 \), we get \( y = 0 \). Substituting \( y = 0 \) into the second equation \( 2xy = 0 \), this holds true for any \( x \). Thus, the possible critical points are along the \( x \)-axis where \( y = 0 \).

### Step 2: Boundary points (constraint \( x^2 + y^2 = 3 \))
Next, we examine the boundary of the set \( D \), which is the portion of the circle \( x^2 + y^2 = 3 \) in the first quadrant.

We parametrize the boundary using polar coordinates:
\[
x = \sqrt{3} \cos \theta, \quad y = \sqrt{3} \sin \theta
\]
where \( 0 \leq \theta \leq \frac{\pi}{2} \) because we are in the first quadrant.

Substitute these into \( f(x, y) \):

\[
f(\sqrt{3} \cos \theta, \sqrt{3} \sin \theta) = (\sqrt{3} \cos \theta)(\sqrt{3} \sin \theta)^2 + 9
\]
\[
= \sqrt{3} \cos \theta (3 \sin^2 \theta) + 9
\]
\[
= 3\sqrt{3} \cos \theta \sin^2 \theta + 9
\]

We now need to maximize and minimize this expression over \( 0 \leq \theta \leq \frac{\pi}{2} \).

### Step 3: Evaluate at critical points and boundary

#### At \( (x, y) = (0, 0) \):
\[
f(0, 0) = 0 \cdot 0^2 + 9 = 9
\]

#### At boundary points:
Let's evaluate the function at a few key points on the boundary.

- **At \( (x, y) = (0, \sqrt{3}) \)** (corresponding to \( \theta = \frac{\pi}{2} \)):
  \[
  f(0, \sqrt{3}) = 0 \cdot (\sqrt{3})^2 + 9 = 9
  \]

- **At \( (x, y) = (\sqrt{3}, 0) \)** (corresponding to \( \theta = 0 \)):
  \[
  f(\sqrt{3}, 0) = (\sqrt{3}) \cdot 0^2 + 9 = 9
  \]

- **At \( (x, y) = (\sqrt{3}/2, \sqrt{3}/2) \)** (corresponding to \( \theta = \frac{\pi}{4} \)):
  \[
  f\left( \frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2} \right) = \frac{\sqrt{3}}{2} \cdot \left( \frac{\sqrt{3}}{2} \right)^2 + 9 = \frac{\sqrt{3}}{2} \cdot \frac{3}{4} + 9 = \frac{3\sqrt{3}}{8} + 9
  \approx 9.649
  \]

### Step 4: Conclusion

Thus, the absolute maximum value is approximately \( 9.649 \), which occurs at \( (x, y) = (\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}) \). The absolute minimum value is \( 9 \), occurring at \( (x, y) = (0, 0) \), \( (0, \sqrt{3}) \), and \( (\sqrt{3}, 0) \).

#### Do you want more details or have any questions?

### 5 Related Questions:
1. How can we apply the Lagrange multiplier method to this problem?
2. How would the solution change if the constraint set \( D \) covered the entire circle?
3. What happens if the function \( f(x, y) \) is changed to \( f(x, y) = x^2 + y^2 \)?
4. How do we find the global maximum of a function over an unbounded region?
5. Can the critical points of \( f(x, y) \) give local extrema even outside the boundary?

### Tip:
When solving optimization problems with constraints, always check both the interior and the boundary for potential maxima and minima.

Find the absolute maximum and minimum values of f on the set D.
f(x, y) = xy^2 + 9,    D = {(x, y) | x ≥ 0, y ≥ 0, x^2 + y^2 ≤ 3}. Give your answer in terms of exact numbers.

This problem explores how to find the absolute maximum and minimum values of the function f(x, y) = xy^2 + 9 within a quarter-circle region in the first quadrant. Learn how to use partial derivatives, polar coordinates, and boundary constraints for constrained optimization. Suitable for students in Grades 11-12 or undergraduate courses in calculus.

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