3. In all other cases - 0 points

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

## Assignment 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

$$
4 x_{1}-3 p_{1}-44=0
$$

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

$$
p_{2}^{2}-12 p_{2}+x_{2}^{2}-8 x_{2}+43=0
$$

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) is the deal 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 $4 x_{1}-3 p_{1}-44=0 \Leftrightarrow x=\frac{3}{4} p+11$ defines the equation of a certain line $l$ in the plane $x O p$. Consider the expression

$$
\begin{aligned}
p^{2}-12 p+x^{2}-8 x+4 & =0 \Leftrightarrow(p-6)^{2}+(x-4)^{2}-36-16+43=0 \\
\Leftrightarrow & (p-6)^{2}+(x-4)^{2}=3^{2}
\end{aligned}
$$

It defines in the plane $x O p$ the equation of a circle $\Phi$ with center at the point $(6 ; 4)$ and radius 3. 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's condition, we need to find the minimum distance between points lying on

![](https://cdn.mathpix.com/cropped/2024_05_06_1805ec3a4f3def5f099cg-13.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):

$$
O_{1} H=\frac{|4 \cdot 4-3 \cdot 6-44|}{\sqrt{3^{2}+4^{2}}}=\frac{46}{5}=9.2 .
$$

Therefore, $H P=9.2-3=6.2$.

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 $O_{1}(6 ; 4)$, its equation in the plane $x O p$ has the form

$$
x=-\frac{4}{3}(p-6)+4 \Leftrightarrow x=-\frac{4}{3} p+12
$$

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:

$$
-\frac{4}{3} p+12=\frac{3}{4} p+11 \Leftrightarrow \mathrm{p}=\frac{12}{25}=0.48
$$

Thus, $p_{1}=0.48$ and $x_{1}=\frac{3}{4} p_{1}+11=11.36$.

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

$$
(p-6)^{2}+\left(-\frac{4}{3} p+12-4\right)^{2}=3^{2}
$$

We get two solutions $p=\frac{39}{5}$ and $p=\frac{21}{5}$. Since the point $P$ lies to the left of the point $O_{1}$, we finally have $p_{2}=\frac{21}{5}=4.2$. Thus, $x_{2}=6.4$.

The advantage of the deal will be:

$$
x_{1}+x_{2}-p_{1}-p_{2}=11.36+6.4-0.48-4.2=13.08
$$

Thus, the deal is profitable for Crocodile Gena, and the profit will be 13080 rubles.

Answer: 1) 6.2. 2) profitable, profit 13080 rubles.

## Criteria: