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 $\theta = \frac{\pi}{4}$):

  \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.