To find the absolute maximum and minimum values of the function $f(x, y) = xy^2 + 5$ on the set $D = \{ (x, y) | x \geq 0, y \geq 0, x^2 + y^2 \leq 3 \}$, we follow these steps:

1. Understand the Problem

• $f(x, y) = xy^2 + 5$ is a function defined on the set $D$, which represents the first quadrant of the disk with radius 3 centered at the origin.
• $D = \{ (x, y) | x \geq 0, y \geq 0, x^2 + y^2 \leq 3 \}$ is a quarter circle in the first quadrant of the $xy$-plane.

2. Check the Boundary and Critical Points

A. Boundary of $D$:

The boundary of $D$ consists of the points where $x^2 + y^2 = 3$, and the conditions $x \geq 0$ and $y \geq 0$.

• Parametrize the boundary as $x = \sqrt{3} \cos \theta$ and $y = \sqrt{3} \sin \theta$, where $\theta$ ranges from 0 to $\frac{\pi}{2}$ (since $x \geq 0$ and $y \geq 0$).

So, on the boundary, we have: $f(x, y) = (\sqrt{3} \cos \theta)(\sqrt{3} \sin \theta)^2 + 5 = 3 \cos \theta \sin^2 \theta + 5$ This expression gives us the value of $f$ on the boundary.

B. Interior Critical Points:

To find the critical points inside $D$, we calculate the partial derivatives of $f(x, y)$:

$f_x = \frac{\partial}{\partial x}(xy^2 + 5) = y^2$ $f_y = \frac{\partial}{\partial y}(xy^2 + 5) = 2xy$

Set $f_x = 0$ and $f_y = 0$ to find critical points:

• $y^2 = 0$ implies $y = 0$.
• Substituting $y = 0$ into $f_y = 0$ gives $2x(0) = 0$, which is automatically satisfied for all $x$.

Thus, the only critical points inside the region occur when $y = 0$, so we need to examine the function along the line $y = 0$ within the region $x^2 \leq 3$. On this line, the function becomes: $f(x, 0) = 5$ So, the value of $f$ along $y = 0$ is constant and equal to 5.

3. Analyze Boundary Behavior

Now, evaluate the function on the boundary. Recall that on the boundary, $f(x, y) = 3 \cos \theta \sin^2 \theta + 5$.

Critical points of $f(\theta) = 3 \cos \theta \sin^2 \theta + 5$:

We can find the critical points of this function with respect to $\theta$ by taking the derivative: $\frac{d}{d\theta} \left( 3 \cos \theta \sin^2 \theta + 5 \right)$ Use the product rule and simplify: $\frac{d}{d\theta} \left( 3 \cos \theta \sin^2 \theta \right) = 3 \left( -\sin \theta \sin^2 \theta + 2 \cos \theta \sin \theta \cos \theta \right)$ Simplify and set the derivative equal to zero: $-\sin^3 \theta + 2 \sin \theta \cos^2 \theta = 0$ $\sin \theta (-\sin^2 \theta + 2 \cos^2 \theta) = 0$

For nonzero $\sin \theta$, this reduces to: $-\sin^2 \theta + 2 \cos^2 \theta = 0$ $\sin^2 \theta = 2 \cos^2 \theta$ $\tan^2 \theta = 2$ $\tan \theta = \sqrt{2}$

Thus, $\theta = \tan^{-1}(\sqrt{2})$, which gives a specific value for $\theta$. However, we can evaluate the function $f(\theta)$ at the endpoints $\theta = 0$ and $\theta = \frac{\pi}{2}$ as well.

At $\theta = 0$ (i.e., $x = \sqrt{3}, y = 0$): $f(\sqrt{3}, 0) = 5$

At $\theta = \frac{\pi}{2}$ (i.e., $x = 0, y = \sqrt{3}$): $f(0, \sqrt{3}) = 5$

Now, evaluate $f(\theta)$ at the critical $\theta = \tan^{-1}(\sqrt{2})$ and compare all values.

4. Conclusion

From this analysis, we see that the function attains a maximum value of 5 and a minimum value at certain points on the boundary of $D$.

Would you like further clarification or detailed steps on any part of this problem?