1. A workshop produces items of types $A$ and $B$. For one item of type $A$, 10 kg of steel and 23 kg of non-ferrous metals are used, and for one item of type $B-70$ kg of steel and 40 kg of non-ferrous metals are used. The profit from selling one item of type $A$ is 80 thousand rubles, and for type $B-100$ thousand rubles. The shift fund for steel is 700 kg, and for non-ferrous metals, it is 642 kg. How many items of types $A$ and $B$ should be produced per shift to maximize the profit from selling the items, given that resource consumption should not exceed the allocated shift funds? Write the maximum profit (in thousands of rubles) that can be obtained under these conditions. Provide the answer as a number without specifying the unit. $\quad$ (5 points)
Solution. Let $x$ be the number of items of type $A$, and $y$ be the number of items of type $B$. Then the profit per shift is calculated by the formula $D=80 x+100 y$, with the constraints $10 x+70 y \leq 700$, $23 x+40 y \leq 642$, and $x$ and $y$ are non-negative integers.

![](https://cdn.mathpix.com/cropped/2024_05_06_fb302ee711d03f314b6cg-09.jpg?height=626&width=1374&top_left_y=932&top_left_x=361)

$10 x+70 y \leq 700 \quad \Leftrightarrow \quad y \leq 10-\frac{x}{7}, \quad 23 x+40 y \leq 642 \quad \Leftrightarrow \quad y \leq 16.05-\frac{23 x}{40}$.

The intersection point of the lines $x+7 y=70$ and $23 x+40 y=642$ is the point with coordinates $x=14, y=8$. In this case, $D=80 \cdot 14+100 \cdot 8=1920$, but when $x=\frac{642}{23}=27 \frac{21}{23}, y=0$, we have the maximum value of the function $D=80 x+100 y$ for non-negative $x$ and $y$ that satisfy the inequalities $10 x+70 y \leq 700, 23 x+40 y \leq 642$. This value is $2233 \frac{1}{23}$. Since $x$ and $y$ are integers, when $y=0, x=27$, the profit $D=2160$; when $y=1, x \leq \frac{602}{23}, x=26, D=2180$; when $y=2, x \leq \frac{562}{23}, x=24, D=2120$. Therefore, the maximum profit is 2180 thousand rubles, given the production of one item of type $A$ and 26 items of type $B$.

## Answer: 2180