3. In all other cases - 0 points

*Note: the numerical assessment of the free area (solution) is not the only possible one, for example, the "gap" can be more than 10 m.

## Task 2 (12 points)

Crocodile Gena and Old Lady Shapoklyak entered into a futures contract, according to which Gena agreed to invest in the 1st project an amount equal to $p_{1}>0$ thousand rubles, and after a year receive $x_{1}$ thousand rubles such that

$$
3 x_{1}-4 p_{1}-30=0
$$

and at the same time Gena agreed to invest in the 2nd project an amount equal to $p_{2}>0$ thousand rubles, and after a year receive $x_{2}$ thousand rubles such that

$$
p_{2}^{2}-12 p_{2}+x_{2}^{2}-14 x_{2}+69=0
$$

According to the terms of the contract, the parameters $p_{1}, x_{1}, p_{2}, x_{2}$ were chosen so that the value

$$
d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$

was minimized. Determine:

1) the minimum value of the quantity $d$;
2) whether the deal is profitable for Gena? Find the amount of profit or loss for Gena if he agrees to this deal.
# Solution:

1) It is clear that the expression defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression

$$
\begin{gathered}
p^{2}-12 p+x^{2}-14 x+69=0 \Leftrightarrow(p-6)^{2}+(x-7)^{2}-36-49+69=0 \\
\Leftrightarrow(p-6)^{2}+(x-7)^{2}=4^{2}
\end{gathered}
$$

It defines in the plane $x O p$ the equation of a circle $\Phi$ with center at the point $(6 ; 7)$ and radius 4. Then the expression of the form

$$
\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(p_{1}-p_{2}\right)^{2}}
$$

defines the distance between points lying on $l$ and $\Phi$. Therefore, according to the problem, we need to find the smallest distance between points lying on

![](https://cdn.mathpix.com/cropped/2024_05_06_1805ec3a4f3def5f099cg-03.jpg?height=609&width=942&top_left_y=375&top_left_x=494)

the line and the circle. Let $O_{1} H$ be the perpendicular from the center of the circle $O_{1}$ to the line $l$, then $O_{1} P$ is the radius of the circle. We will show that $H P=O_{1} H-O_{1} P$ is the desired value. Indeed, let $H_{1} \in l, F \in \Phi$. Draw the tangent $a$ to the circle at point $P$. The radius $O_{1} P$, drawn to the point of tangency, is perpendicular to $a$. Therefore, $a \| l$. And thus, $HP$ is equal to the distance between the parallel lines. Now we have:

$$
H_{1} F=H_{1} P_{1}+P_{1} F \geq H_{1} P_{1} \geq H P .
$$

Thus, $H P$ is the shortest distance between the points of the circle and the line $l$. Let's find $O_{1} H$ using the formula for the distance from a point to a line (one of the ways to find it):

$$
O_{1} H=\frac{|3 \cdot 7-4 \cdot 6-30|}{\sqrt{3^{2}+4^{2}}}=\frac{33}{5}=6.6 \text {. }
$$

Therefore, $H P=6.6-4=2.6$.

2) Let's define the equation of the line $O_{1} H$. Since it is perpendicular to the line $l$ and passes through the point $O_{1}(6 ; 7)$, its equation in the plane $x O p$ has the form

$$
x=-\frac{3}{4}(p-6)+7 \Leftrightarrow x=-\frac{3}{4} p+\frac{23}{2}
$$

Let's find the coordinates of the point $H$ as the coordinates of the intersection of the lines $l$ and $O_{1} H$. Form the equation:

From here, $p_{1}=0.72$ and $x_{1}=\frac{4}{3} p_{1}+10=10.96$.

Let's find the coordinates of the point $P$ as the coordinates of the intersection of the line $O_{1} H$ and the circle 9 :

$$
(p-6)^{2}+\left(-\frac{3}{4} p+\frac{23}{2}-7\right)^{2}=4^{2}
$$

We get two solutions $p=\frac{46}{5}$ and $p=\frac{14}{5}$. Since the point $P$ lies to the left of the point $O_{1}$, we finally have $p_{2}=\frac{14}{5}=2.8$. From here, $x_{2}=9.4$.

The advantage of the deal will be:

$$
x_{1}+x_{2}-p_{1}-p_{2}=10.96+9.4-2.8-0.72=16.84 .
$$

Thus, the deal is profitable for Crocodile Gena, and the profit will be 16840 rubles.

Answer: 1) 2.6. 2) profitable, profit 16840 rubles.