A chicken coop with an area of $240 \mathrm{~m}^{2}$ is to house chickens and chicks, with a desirable free space of $4 \mathrm{~m}^{2}$ for each chicken and $2 \mathrm{~m}^{2}$ for each chick. Additionally, each chick consumes $40 \mathrm{~g}$ of feed per day, and each chicken consumes $160 \mathrm{~g}$ per day, with a maximum daily feed allowance of $8 \mathrm{~kg}$.

(a) Represent the conditions of the problem algebraically.

(b) Graphically represent the conditions above on the Cartesian plane $x O y$.

(c) Can the chicken coop accommodate 20 chickens and 80 chicks? And 30 chickens and 100 chicks?

(d) What is the maximum number of chickens that can be placed in the chicken coop, respecting the desirable spaces and the maximum feed allowance? And of chicks?
Let $x$ and $y$ be, respectively, the number of chickens and chicks in the coop.

(a) We have $4 x+2 y=240$, that is, $2 x+y=120$.

Since $8 \text{ kg} = 8000 \text{ g}$, we have: $160 x+40 y \leq 8000$. Thus, $4 x+y \leq 200$. In summary, the number $x$ of chickens and $y$ of chicks satisfy:

$$
(*)\left\{\begin{array}{l}
2 x+y=120 \\
4 x+y \leq 200
\end{array}\right.
$$

(b) The line $2 x+y=120$ intersects the $O x$ axis at $x=60$ and the $O y$ axis at $y=120$.

The line $4 x+y=200$ intersects the $O x$ axis at $x=50$ and the $O y$ axis at $y=200$. The graphs of these lines are shown below, where the inequality $4 x+y \leq 200$ is represented by the shaded region.
![](https://cdn.mathpix.com/cropped/2024_05_01_a444074f1f8ae50444d5g-12.jpg?height=502&width=1352&top_left_y=1641&top_left_x=343)

Observe that the condition $(*)$ is represented in the figure by the segment connecting the points $P$ and $(0,120)$. The coordinates of the point $P$ are the solution to the system:

$$
\begin{aligned}
& \qquad\left\{\begin{array}{l}
2 x+y=120 \\
4 x+y=200
\end{array}\right. \\
& \text { that is, } x=40 \text { and } y=40, \text { and } \\
& P=(40,40)
\end{aligned}
$$

(c) We have $2 \times 20+80=120$ and $4 \times 20+80 \leq 200$. Therefore, $x=20$ and $y=80$ satisfy the condition $\left({ }^{*}\right)$, and thus the answer is yes.

Now $2 \times 30+100 \neq 120$, so $x=30$ and $y=100$ do not satisfy the condition (*) and, therefore, the answer is no.

(d) The maximum number of chickens is 40, and in this case we will also have 40 chicks. The maximum number of chicks is 120, and in this case we will have 0 chickens.