$$ 1 \leftarrow $$ 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 $$ f(x, y) $$. $$ f_{x}(x, y)=\square $$ and $$ f_{y}(x, y)=\square $$

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The function is given by  
$$
f(x, y)=\sqrt{3x^2+3y^2-6x+7}.
$$
Let's define  
$$
u(x,y)=3x^2+3y^2-6x+7,
$$
so that  
$$
f(x,y)=\sqrt{u(x,y)}.
$$

Using the chain rule:

1. The partial derivative with respect to $$ x $$ is  
$$
f_{x}(x,y)=\frac{1}{2\sqrt{u(x,y)}}\cdot \frac{\partial u}{\partial x}.
$$
Differentiate $$ u(x,y) $$ with respect to $$ x $$:
$$
\frac{\partial u}{\partial x}=6x-6.
$$
Thus,
$$
f_{x}(x,y)=\frac{6x-6}{2\sqrt{3x^2+3y^2-6x+7}}.
$$
This expression can also be factored as:
$$
f_{x}(x,y)=\frac{6(x-1)}{2\sqrt{3x^2+3y^2-6x+7}}=\frac{3(x-1)}{\sqrt{3x^2+3y^2-6x+7}}.
$$

2. The partial derivative with respect to $$ y $$ is  
$$
f_{y}(x,y)=\frac{1}{2\sqrt{u(x,y)}}\cdot \frac{\partial u}{\partial y}.
$$
Differentiate $$ u(x,y) $$ with respect to $$ y $$:
$$
\frac{\partial u}{\partial y}=6y.
$$
Thus,
$$
f_{y}(x,y)=\frac{6y}{2\sqrt{3x^2+3y^2-6x+7}}=\frac{3y}{\sqrt{3x^2+3y^2-6x+7}}.
$$

Therefore, the two partial derivatives are:

$$
f_{x}(x,y)=\frac{3(x-1)}{\sqrt{3x^2+3y^2-6x+7}} \quad \text{and} \quad f_{y}(x,y)=\frac{3y}{\sqrt{3x^2+3y^2-6x+7}}.
$$
$$ f_{x}(x, y) = \frac{3(x - 1)}{\sqrt{3x^2 + 3y^2 - 6x + 7}} $$ and $$ f_{y}(x, y) = \frac{3y}{\sqrt{3x^2 + 3y^2 - 6x + 7}} $$.

Find the absolute maximum and minimum values of the following function on the given region .

Find both partial derivatives of .
and

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