Find the absolute maximum and minimum values of the following function on the given region $$ R $$. $$ f(x, y)=\sqrt{3 x^{2}+3 y^{2}-6 x+7} ; R=\left\{(x, y): x^{2}+y^{2} \leq 4, y \geq 0\right\} $$ Find both partial derivatives of $$ f(x, y) $$. $$ f_{x}(x, y)=\frac{3(x-1)}{\sqrt{3 x^{2}+3 y^{2}-6 x+7}} $$ and $$ f_{y}(x, y)=\frac{3 y}{\sqrt{3 x^{2}+3 y^{2}-6 x+7}} $$ Determine the absolute maximum value of $$ f $$ on $$ R $$. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. There is no absolute maximum value. B. The absolute maximum value of $$ f $$ on $$ R $$ is (Type an exact answer.)

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To find the absolute maximum and minimum values of the function $$ f(x, y) = \sqrt{3x^2 + 3y^2 - 6x + 7} $$ on the region $$ R = \{(x, y): x^2 + y^2 \leq 4, y \geq 0\} $$, we will follow these steps:

1. **Find the critical points of $$ f $$ in the interior of the region $$ R $$** by setting the partial derivatives equal to zero.
2. **Evaluate $$ f $$ on the boundary of the region $$ R $$**.
3. **Compare the values from the critical points and the boundary to determine the absolute maximum and minimum.**

### Step 1: Find Critical Points

The partial derivatives are given as:
$$
f_{x}(x, y) = \frac{3(x-1)}{\sqrt{3x^{2}+3y^{2}-6x+7}}
$$
$$
f_{y}(x, y) = \frac{3y}{\sqrt{3x^{2}+3y^{2}-6x+7}}
$$

Setting $$ f_{x}(x, y) = 0 $$ gives:
$$
3(x-1) = 0 \implies x = 1
$$

Setting $$ f_{y}(x, y) = 0 $$ gives:
$$
3y = 0 \implies y = 0
$$

Thus, the critical point is $$ (1, 0) $$.

### Step 2: Evaluate $$ f $$ on the Boundary

The boundary of the region $$ R $$ is defined by $$ x^2 + y^2 = 4 $$ and $$ y \geq 0 $$. We can parameterize the boundary using polar coordinates:
$$
x = 2\cos(\theta), \quad y = 2\sin(\theta) \quad \text{for } 0 \leq \theta \leq \pi
$$

Substituting into $$ f $$:
$$
f(2\cos(\theta), 2\sin(\theta)) = \sqrt{3(2\cos(\theta))^2 + 3(2\sin(\theta))^2 - 6(2\cos(\theta)) + 7}
$$
$$
= \sqrt{12 - 12\cos(\theta) + 7} = \sqrt{19 - 12\cos(\theta)}
$$

Now we need to find the maximum and minimum values of $$ \sqrt{19 - 12\cos(\theta)} $$ for $$ 0 \leq \theta \leq \pi $$.

### Step 3: Analyze the Function on the Boundary

The function $$ \cos(\theta) $$ ranges from $$ 1 $$ to $$ -1 $$ as $$ \theta $$ goes from $$ 0 $$ to $$ \pi $$:
- At $$ \theta = 0 $$: $$ f(2, 0) = \sqrt{19 - 12(1)} = \sqrt{7} $$
- At $$ \theta = \pi $$: $$ f(-2, 0) = \sqrt{19 - 12(-1)} = \sqrt{31} $$

The maximum occurs at $$ \theta = \pi $$ and the minimum occurs at $$ \theta = 0 $$.

### Step 4: Compare Values

Now we compare the values at the critical point and the boundary:
- At the critical point $$ (1, 0) $$:
$$
f(1, 0) = \sqrt{3(1)^2 + 3(0)^2 - 6(1) + 7} = \sqrt{4} = 2
$$
- At the boundary:
  - $$ f(2, 0) = \sqrt{7} \approx 2.64575 $$
  - $$ f(-2, 0) = \sqrt{31} \approx 5.56776 $$

### Conclusion

The absolute maximum value of $$ f $$ on $$ R $$ is $$ \sqrt{31} $$.

Thus, the answer is:
B. The absolute maximum value of $$ f $$ on $$ R $$ is $$ \sqrt{31} $$.
The absolute maximum value of $$ f $$ on $$ R $$ is $$ \sqrt{31} $$.

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