A chicken coop with $240 \mathrm{~m}^{2}$ of area must 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 expenditure of $8 \mathrm{~kg}$.

(a) Represent the conditions of the problem algebraically.

(b) Represent graphically, on the Cartesian plane $x O y$, the conditions of the problem.

(c) Can this 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 expenditure? And of chicks?
Let $x$ and $y$ be, respectively, the number of hens and chicks in the henhouse.

(a) We have $4 x+2 y=240$, which simplifies to $2 x+y=120$. Since $8 \text{ kg} = 8000 \text{ g}$, we have $160 x+40 y \leq 8000$. Thus, $4 x+y \leq 200$.

In summary, the numbers $x$ of hens and $y$ of chicks satisfy

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

![](https://cdn.mathpix.com/cropped/2024_05_01_30c9a294a58a6e4b8190g-100.jpg?height=422&width=734&top_left_y=931&top_left_x=1095)

(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 in the figure, where the inequality $4 x+y \leq 200$ is represented by the shaded region. Observe that, in the figure, the conditions $(*)$ are represented by the segment connecting the points $P$ and $(0,120)$. The coordinates of the point $P$ are the solution to the system

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

which gives $x=40$ and $y=40$, so $P=(40,40)$.

(c) We have $2 \times 20 + 80 = 120$ and $4 \times 20 + 80 \leq 200$. Therefore, $x=20$ and $y=80$ satisfy the conditions $(*)$ and, thus, the henhouse can indeed accommodate 20 hens and 80 chicks. Now, $2 \times 30 + 100 \neq 120$, so $x=30$ and $y=100$ do not satisfy the conditions $(*)$ and, thus, the henhouse cannot accommodate 30 hens and 100 chicks.

(d) The maximum number of hens 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 hens.