2. For numbers $a$ and $b$ such that $a + 10b = 1$, what is the maximum possible value of the product $a \cdot b$? Determine the values of $a$ and $b$ for which this value is achieved.
2. For numbers $a$ and $b$ such that $a+10 b=1$, what is the maximum possible value of the product $a \cdot b$? Determine the values of $a$ and $b$ for which this value is achieved.

## Solution.

$$
\begin{aligned}
a b & =(1-10 b) b=b-10 b^{2} \\
& =-10 b^{2}+b=-10\left(b^{2}-\frac{1}{10} b\right) \\
& =-10\left(b^{2}-\frac{1}{10} b+\frac{1}{400}-\frac{1}{400}\right) \\
& =-10\left(b^{2}-\frac{1}{10} b+\frac{1}{400}\right)+\frac{1}{40} \\
& =\frac{1}{40}-10\left(b-\frac{1}{20}\right)^{2} .
\end{aligned}
$$

The square of any number is a non-negative number, and so is its tenfold.

Therefore, it follows that

$\frac{1}{40}-10\left(b-\frac{1}{20}\right)^{2} \leq \frac{1}{40}$.

If $a b=\frac{1}{40}$, then

$10\left(b-\frac{1}{20}\right)^{2}=0$

$b-\frac{1}{20}=0$

$b=\frac{1}{20}$.

Finally, we have

$a=1-10 b=1-10 \cdot \frac{1}{20}=\frac{1}{2}$.

Thus, the maximum possible value of the product $a \cdot b$ is $\frac{1}{40}$.