Let's determine the maximum value of the expression

$$
\sqrt{x-2}+2 \sqrt{3-x}
$$
I. solution. The domain of the expression: $2 \leq x \leq 3$.

Apply the inequality between the arithmetic and quadratic means:

$$
\begin{aligned}
& \frac{\sqrt{x-2}+\frac{1}{2} \sqrt{3-x}+\frac{1}{2} \sqrt{3-x}+\frac{1}{2} \sqrt{3-x}+\frac{1}{2} \sqrt{3-x}}{5} \leq \sqrt{\frac{x-2+\frac{1}{4}(3-x) \cdot 4}{5}} \\
& \frac{\sqrt{x-2}+2 \sqrt{3-x}}{5} \leq \sqrt{\frac{1}{5}}
\end{aligned}
$$

Multiplying both sides by 5:

$$
\sqrt{x-2}+2 \sqrt{3-x} \leq \sqrt{5}
$$

Equality holds if and only if $\sqrt{x-2}=\frac{1}{2} \sqrt{3-x}$. The solution to this equation is $x=\frac{11}{5}$, which is within the domain of the expression.

Thus, the expression attains its maximum at $x=\frac{11}{5}$, and the value of the maximum is $\sqrt{5}$.

II. solution. Let

$$
f(x)=\sqrt{x-2}+2 \sqrt{3-x}=\sqrt{x-2}+\sqrt{12-4 x}
$$

The function is defined only on the interval $2 \leq x \leq 3$.

![](https://cdn.mathpix.com/cropped/2024_05_02_b4b590f3f985a9047131g-1.jpg?height=399&width=436&top_left_y=1042&top_left_x=839)

To determine the extremum, we differentiate the function. Since differentiation is only possible on an open interval, we examine the expression on the interval $(2, 3)$.

| $x$ | $2$ | $2.2$ | $3$ |
|-----|-----|-------|-----|
| $f'(x)$ | $>0$ | $0$ | $<0$ |
| $f$ | $\nearrow$ | $\sqrt{5}$ | $\searrow$ |

Thus, the function has a maximum at $x=2.2$ on the interval $(2, 3)$.

We must also check the values of the expression at $x=2$ and $x=3$, where the values are 2 and 1, respectively. Since both are less than $\sqrt{5}$, the maximum value of the function $f$ is at $x=2.2$, and the maximum value is $\sqrt{5}$.

Note. Many people forget to check the values of $f$ at the endpoints of the interval $[2, 3]$.

III. solution. The domain of $\sqrt{x-2}+2 \sqrt{3-x}$ is $[2, 3]$. Notice that

$$
(\sqrt{x-2})^{2}+(\sqrt{3-x})^{2}=x-2+3-x=1
$$

This means there is an acute angle $\beta$ such that $\sqrt{x-2}=\sin \beta$ and $\sqrt{3-x}=\cos \beta$, so

$$
\sqrt{x-2}+2 \sqrt{3-x}=\sin \beta+2 \cos \beta=\sqrt{5}\left(\frac{1}{\sqrt{5}} \sin \beta+\frac{2}{\sqrt{5}} \cos \beta\right)
$$

Since $\left(\frac{1}{\sqrt{5}}\right)^{2}+\left(\frac{2}{\sqrt{5}}\right)^{2}=1$, there is an acute angle $\alpha$ such that $\frac{1}{\sqrt{5}}=\cos \alpha$ and $\frac{2}{\sqrt{5}}=\sin \alpha$. Using this, the expression can be written as:

$$
\sqrt{5}(\cos \alpha \sin \beta+\sin \alpha \cos \beta)=\sqrt{5} \sin (\alpha+\beta)
$$

The sine function attains its maximum value of 1 for acute angles when $\alpha+\beta=90^{\circ}$, i.e., $\beta=90^{\circ}-\alpha$. Then

$$
\sqrt{x-2}=\sin \beta=\sin \left(90^{\circ}-\alpha\right)=\cos \alpha=\frac{1}{\sqrt{5}}
$$

From this, we can calculate the value of $x$: $\sqrt{x-2}=\frac{1}{\sqrt{5}}$, so $x-2=\frac{1}{5}$, and thus $x=2+\frac{1}{5}$.

The maximum value of the expression is:

$$
\sqrt{2+\frac{1}{5}-2}+2 \sqrt{3-2-\frac{1}{5}}=\sqrt{\frac{1}{5}}+2 \sqrt{\frac{4}{5}}=\frac{1}{\sqrt{5}}+\frac{4}{\sqrt{5}}=\frac{5}{\sqrt{5}}=\sqrt{5}
$$

IV. solution. Introduce the following vectors:

$$
\mathbf{a}=(1, 2), \quad \mathbf{b}=(\sqrt{x-2}, \sqrt{3-x})
$$

Let $\gamma$ be the angle between the two vectors. Write the dot product of the two vectors in two ways:

$$
\mathbf{a} \cdot \mathbf{b}=|\mathbf{a}| \cdot|\mathbf{b}| \cdot \cos \gamma=1 \cdot \sqrt{x-2}+2 \cdot \sqrt{3-x}
$$

Since $\cos \gamma \leq 1$, we have $|\mathbf{a}| \cdot|\mathbf{b}| \geq 1 \cdot \sqrt{x-2}+2 \cdot \sqrt{3-x}$. Since $|\mathbf{a}|=\sqrt{1+4}=\sqrt{5}$ and $|\mathbf{b}|=\sqrt{x-2+3-x}=1$, we get

$$
\sqrt{5} \cdot 1 \geq \sqrt{x-2}+2 \cdot \sqrt{3-x}
$$

Thus, the maximum value of the expression is $\sqrt{5}$.

We need to calculate for which $x$ this value is attained.

$$
\sqrt{x-2}+2 \cdot \sqrt{3-x}=\sqrt{5}
$$

Squaring both sides and simplifying:

$$
3 x-5=4 \cdot \sqrt{x-2} \cdot \sqrt{3-x}
$$

Squaring again and simplifying:

$$
0=25 x^{2}-110 x+121=(5 x-11)^{2}
$$

Thus, the expression attains its maximum value at $x=\frac{11}{5}$.

V. solution. The domain of the expression $\sqrt{x-2}+2 \sqrt{3-x}$ is $2 \leq x \leq 3$. By plotting in Excel, it can be inferred that the function attains its maximum at $x=2.2$. (We calculated the expression for a thousand values from 2 to 3 with a step size of 0.001, and from the plotted graph, it can be inferred that the maximum is around $x=2.2$, which we also verified by checking the values in the table.)

We need to show that $\sqrt{x-2}+2 \sqrt{3-x} \leq \sqrt{0.2}+2 \sqrt{0.8}$. Both sides are non-negative, so we can square them:

$$
\begin{gathered}
x-2+12-4 x+4 \sqrt{x-2} \cdot \sqrt{3-x} \leq 0.2+3.2+4 \sqrt{0.2} \cdot \sqrt{0.8} \\
4 \sqrt{x-2} \cdot \sqrt{3-x} \leq-5+3 x
\end{gathered}
$$

The left side is positive and for $x \geq 2$ the right side is also positive, so we can square again:

$$
\begin{gathered}
16 \cdot\left(3 x+2 x-x^{2}-6\right) \leq 9 x^{2}-30 x+25 \\
0 \leq 25 x^{2}-110 x+121=(5 x-11)^{2}
\end{gathered}
$$

This is always true, and the steps can be reversed (due to the remarks made along the way).

Thus, we have shown that the expression indeed attains its maximum at $x=2.2$, and the value of the maximum is $\sqrt{0.2}+2 \sqrt{0.8} \approx 2.236$.