## A1 MLD

Let $x, y, z$ be real numbers, satisfying the relations

$$
\left\{\begin{array}{l}
x \geq 20 \\
y \geq 40 \\
z \geq 1675 \\
x+y+z=2015
\end{array}\right.
$$

Find the greatest value of the product $P=x \cdot y \cdot z$.


## Solution 1:

By virtue of $z \geq 1675$ we have

$$
y+z<2015 \Leftrightarrow y<2015-z \leq 2015-1675<1675
$$

It follows that $(1675-y) \cdot(1675-z) \leq 0 \Leftrightarrow y \cdot z \leq 1675 \cdot(y+z-1675)$.

By using the inequality $u \cdot v \leq\left(\frac{u+v}{2}\right)^{2}$ for all real numbers $u, v$ we obtain

$$
\begin{gathered}
P=x \cdot y \cdot z \leq 1675 \cdot x \cdot(y+z-1675) \leq 1675 \cdot\left(\frac{x+y+z-1675}{2}\right)^{2}= \\
1675 \cdot\left(\frac{2015-1675}{2}\right)^{2}=1675 \cdot 170^{2}=48407500
\end{gathered}
$$

$$
\text { We have } P=x \cdot y \cdot z=48407500 \Leftrightarrow\left\{\begin{array} { l } 
{ x + y + z = 2 0 1 5 , } \\
{ z = 1 6 7 5 , } \\
{ x = y + z - 1 6 7 5 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x=170 \\
y=170 \\
z=1675
\end{array}\right.\right.
$$

So, the greatest value of the product is $P=x \cdot y \cdot z=48407500$.

