1. A workshop produces transformers of types $A$ and $B$. For one transformer of type $A$, 5 kg of transformer iron and 3 kg of wire are used, and for a transformer of type $B$, 3 kg of iron and 2 kg of wire are used. The profit from selling a transformer of type $A$ is 12 thousand rubles, and for type $B$ it is 10 thousand rubles. The shift fund of iron is 481 kg, and wire is 301 kg. How many transformers of types $A$ and $B$ should be produced per shift to maximize the profit from selling the products, if the resource consumption should not exceed the allocated shift funds? What will be the maximum profit in this case?

$(12$ points)
Solution. Let $x$ be the number of transformers of type $A$, and $y$ be the number of transformers of type $B$. Then the profit per shift is calculated by the formula $D=12 x+10 y$, with the conditions

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f8bb21871417ffa2deg-5.jpg?height=651&width=1099&top_left_y=877&top_left_x=433)

$5 x+3 y \leq 481, \quad 3 x+2 y \leq 301, x$ and $y$ - non-negative integers.

$5 x+3 y \leq 481 \Leftrightarrow y \leq 160 \frac{1}{3}-\frac{5 x}{3}, \quad 3 x+2 y \leq 301 \Leftrightarrow y \leq 150.5-\frac{3 x}{2}$. The intersection point of the lines $5 x+3 y=481$ and $3 x+2 y=301$ is the point with coordinates $x=59, y=62$. In this case, $D=12 \cdot 59+10 \cdot 62=1328$, but for $x=0, y=150.5$ we have the maximum value of the function $D=12 x+10 y$ for non-negative $x$ and $y$ satisfying the inequalities $5 x+3 y \leq 481, 3 x+2 y \leq 301$. This value is 1505. Since $x$ and $y$ are integers, for $x=0, y=150$ the revenue $D=1500$, for $x=1, y \leq 150.5-1.5, y=149, D=1502$, for $x=2, y \leq 150.5-3, y=147, D<1502$.

Thus, the maximum revenue is 1502 thousand rubles, provided that one transformer of type $A$ and 149 transformers of type $B$ are produced.

Answer: 1 transformer of type $A$ and 149 transformers of type $B$, 1502 thousand rubles.

Solution. $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$,

$\left(1+(\sin (x+y)+1)^{2}\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$. Since $1+(\sin (x+y)+1)^{2} \geq 1$, and $3^{x}+3^{-x} \geq 2$, $\log _{2}\left(3^{x}+3^{-x}\right) \geq 1$ for any values of $x$ and $y$, then $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \geq 1$, and the inequality $\left(2+\sin ^{2}(x+y)+2 \sin (x+y)\right) \log _{2}\left(3^{x}+3^{-x}\right) \leq 1$ is valid only for those $x$ and $y$ for which $\left\{\begin{array}{c}\sin (x+y)=-1, \\ 3^{x}+3^{-x}=2,\end{array}\left\{\begin{array}{c}y=-\pi / 2+2 \pi n, n \in N, \\ x=0 .\end{array}\right.\right.$

Answer: $x=0, y=-\pi / 2+2 \pi n, n \in N$.