# Solution:

Let the annual profit $f(x)=p x-q=p x-0.5 x^{2}+2 x+10=$

$-0.5 x^{2}+x(p+2)+10=-0.5\left(x^{2}-2 x(p+2)-20\right)=$

$=-0.5\left(x^{2}-2 x(p+2)+(p+2)^{2}-(p+2)^{2}-20\right)=$

$=-0.5(x-(p+2))^{2}+(p+2)^{2} / 2+10$

The quadratic trinomial $\mathrm{f}(\mathrm{x})$ reaches its maximum value at $\mathrm{x}=\mathrm{p}+2$ and is equal to $(p+2)^{2} / 2+10$

In three years, the total profit should be at least 126 million rubles, so we get

$(p+2)^{2} / 2+10 \geq 126 / 3$

$(p+2)^{2} \geq 42 \cdot 2-20$

![](https://cdn.mathpix.com/cropped/2024_05_06_affb3fefe05593c2a003g-4.jpg?height=120&width=866&top_left_y=1371&top_left_x=298)

but $\mathrm{p}>0=>p \geq 6$.

Answer: $\mathrm{p}=6$.
7. The production of x thousand units of a product costs $q=0.5 x^{2}-2 x-10$ million rubles per year. At a price of p thousand rubles per unit, the annual profit from selling this

Preliminary (online) stage of the "Step into the Future" School Students' Olympiad in the subject of Mathematics

product (in million rubles) is $p x - q$. The factory produces the product in such a quantity that the profit is maximized. For what minimum value of p will the total profit over three years be at least 126 million rubles.