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

Find both partial derivatives of .
and
Determine the absolute maximum value of on . 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 on is .
(Type an exact answer.)
Determine the absolute minimum value of on . Select the correct choice below and, if necessary, fill in the answer box
to complete your choice.
A. The absolute minimum value of on is
(Type an exact answer.)
B. There is no absolute minimum value.

Ask by Joseph Carter. in the United States

Mar 21,2025

Question

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

Find both partial derivatives of .
and
Determine the absolute maximum value of on . 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 on is .
(Type an exact answer.)
Determine the absolute minimum value of on . Select the correct choice below and, if necessary, fill in the answer box
to complete your choice.
A. The absolute minimum value of on is
(Type an exact answer.)
B. There is no absolute minimum value.

Ask by Joseph Carter. in the United States

Mar 21,2025

UpStudy AI · Accepted Answer

$$
\textbf{Step 1. Rewrite the Function by Completing the Square}
$$

We are given
$$
f(x,y)=\sqrt{3x^{2}+3y^{2}-6x+7}.
$$
Factor out the 3 from the terms involving $$x$$ and $$y$$:
$$
f(x,y)=\sqrt{3(x^{2}+y^{2}-2x)+7}.
$$
Complete the square in $$x$$. Notice:
$$
x^2 - 2x = (x-1)^2 - 1.
$$
Hence,
$$
x^{2}+y^{2}-2x = (x-1)^2+y^{2}-1.
$$
Substitute back into the expression for $$f(x,y)$$:
$$
f(x,y)=\sqrt{3\Bigl((x-1)^2+y^2-1\Bigr)+7}=\sqrt{3((x-1)^2+y^2)+4}.
$$

$$
\textbf{Step 2. Find the Partial Derivatives of $$ f(x,y) $$}
$$

We use the chain rule. Let
$$
u(x,y)=3((x-1)^2+y^2)+4,
$$
so that
$$
f(x,y)=\sqrt{u(x,y)}.
$$

The partial derivatives are given by:
$$
f_{x}(x,y)=\frac{1}{2\sqrt{u(x,y)}}\cdot u_{x}(x,y)
\quad \text{and} \quad
f_{y}(x,y)=\frac{1}{2\sqrt{u(x,y)}}\cdot u_{y}(x,y).
$$

First, compute $$u_{x}(x,y)$$:
$$
u(x,y)=3((x-1)^2+y^2)+4,
$$
so
$$
\frac{\partial}{\partial x} ((x-1)^2)=2(x-1) \quad \Longrightarrow \quad u_{x}(x,y)=3\cdot 2(x-1)=6(x-1).
$$

Next, compute $$u_{y}(x,y)$$:
$$
\frac{\partial}{\partial y}(y^2)=2y \quad \Longrightarrow \quad u_{y}(x,y)=3\cdot 2y=6y.
$$

Thus, the partial derivatives are:
$$
f_{x}(x,y)=\frac{6(x-1)}{2\sqrt{3((x-1)^2+y^2)+4}}=\frac{3(x-1)}{\sqrt{3x^2+3y^2-6x+7}},
$$
$$
f_{y}(x,y)=\frac{6y}{2\sqrt{3((x-1)^2+y^2)+4}}=\frac{3y}{\sqrt{3x^2+3y^2-6x+7}}.
$$

$$
\textbf{Step 3. Analyze the Function on the Region $$ R $$}
$$

The region is given by
$$
R=\{(x,y): x^2+y^2\le 4,\; y\ge 0\},
$$
which is the closed upper half of a disk centered at $$(0,0)$$ with radius 2.

Since
$$
f(x,y)=\sqrt{3((x-1)^2+y^2)+4},
$$
the function inside the square root is
$$
3((x-1)^2+y^2)+4.
$$
The minimum and maximum of $$ f $$ on $$ R $$ occur when the term $$(x-1)^2+y^2$$ is minimized and maximized, respectively.

$$
\textbf{Step 4. Determine the Absolute Minimum Value}
$$

The expression $$(x-1)^2+y^2$$ represents the squared distance from the point $$(1,0)$$. To minimize this distance over $$ R $$, choose the point in $$ R $$ closest to $$(1,0)$$.

Clearly, $$(1,0)$$ itself is in $$ R $$ because:
$$
1^2+0^2=1\le 4 \quad \text{and} \quad y=0\ge 0.
$$
At $$(1,0)$$:
$$
(x-1)^2+y^2 = 0.
$$
Thus,
$$
f(1,0)=\sqrt{3\cdot0+4}=\sqrt{4}=2.
$$
So, the absolute minimum value of $$ f $$ on $$ R $$ is $$2$$.

$$
\textbf{Step 5. Determine the Absolute Maximum Value}
$$

To maximize $$ f $$, we maximize $$(x-1)^2+y^2$$ over $$ R $$. Note that on the boundary $$ x^2+y^2=4 $$, we can rewrite:
$$
(x-1)^2+y^2 = x^2-2x+1+y^2 = (x^2+y^2)-2x+1 = 4-2x+1=5-2x.
$$
This expression is maximized when $$ x $$ is minimized. Under $$ x^2+y^2=4 $$ with $$ y\ge 0 $$, the smallest possible $$ x $$ is $$ x=-2 $$ (when $$ y=0 $$).

At $$(-2,0)$$:
$$
(x-1)^2+y^2=(-2-1)^2+0^2= (-3)^2=9.
$$
Thus,
$$
f(-2,0)=\sqrt{3\cdot 9+4}=\sqrt{27+4}=\sqrt{31}.
$$
So, the absolute maximum value of $$ f $$ on $$ R $$ is $$\sqrt{31}$$.

$$
\textbf{Final Answers:}
$$
- The absolute maximum value of $$ f $$ on $$ R $$ is $$\sqrt{31}$$.

- The absolute minimum value of $$ f $$ on $$ R $$ is $$2$$.
The absolute maximum value of $$ f $$ on $$ R $$ is $$ \sqrt{31} $$, and the absolute minimum value is $$ 2 $$.

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