# Problem 4. Maximum 20 points

Svetlana is convinced that any feline can be interesting to her for two purposes - catching mice and conducting cat therapy. Based on these considerations, Svetlana has decided to acquire 2 kittens, each of which has an equal probability of being either a male or a female. The weekly production possibility curve for males is $\mathrm{M}=80-4 \mathrm{~K}$, and for females $\mathrm{M}=16-0.25 \mathrm{~K}$, where $\mathrm{M}$ is the number of mice caught per week, and $\mathrm{K}$ is the number of hours devoted to cat therapy.

(a) Determine the maximum number of mice that 2 kittens can catch in a week with the optimal combination of males and females for this purpose.

(b) Determine the number of different variants of the combined production possibility curve Svetlana has when acquiring 2 kittens.

(c) Determine the analytical and graphical form of the combined production possibility curve for 2 kittens for each of the possible combinations of males and females.

(d) In the end, the kitten seller delivered the order, consisting of a male and a female, but also noted that the male has a twin brother, whom Svetlana can receive for free in her caring hands. Determine the analytical and graphical form of the combined production possibility curve if Svetlana agrees to the offer to take the third kitten.
# Solution:

(a) Males catch mice better than females. Two males can catch $40 \cdot 2=80$ mice in a week.

Answer: 160 (2 points).

(b) There are 3 options: one male and one female, two males, two females.

Answer: 3 (2 points).

(c) Two males with individual PPFs: $M=80-4K$. The opportunity costs for males are constant and equal (they can produce cat therapy in any order). The first produces $K_1$ units of cat therapy, and the second produces $K_2$ units of cat therapy. $K_{1}+K_{2}=K$. Joint PPF: $M=80-4 K_{1}+80-4 K_{2}$.

$$
M=80-4\left(K-K_{2}\right)+80-4 K_{2}=160-4 K
$$

Answer: two males $M=160-4 K, K \leq 40+$ graph (2 points).

Two females with individual PPFs: $M=16-0.25 K$. The solution is similar to the previous part: $M=32-0.25 K, K \leq 128$.

Answer: two females $M=32-0.25 K, K \leq 128+$ graph (2 points).

One male $M_{1}=80-4 K_{1}$ and one female $M_{2}=16-0.25 K_{2}$. The opportunity costs for the female are lower, so she produces cat therapy first.

$$
\begin{gathered}
K_{1}=0, K_{2}=K, K \leq 64 \\
K_{1}=K-64, K_{2}=64,64<K \leq 84
\end{gathered}
$$

The joint PPF can be obtained by substituting these values into the function $M=M_1 + M_2 = 80-4 K_{1} + 16-0.25 K_{2}$:

PPF:

$$
\begin{aligned}
& M=96-0.25 K, 0<K \leq 64 \\
& M=336-4 K, 64<K \leq 84
\end{aligned}
$$

Answer:

$$
M=96-0.25 K, 0<K \leq 64
$$

$M=336-4 K, 64<K \leq 84+$ graph (4 points)

(d) The problem reduces to finding the joint PPF for three individual functions:

$M=80-2 K, M=80-2 K, M=16-0.25 K$. Clearly, the problem can be simplified by combining the joint PPF of two males from the previous part $M=160-4 K$ with the PPF of one female $M=16-0.25 K$.

Similarly to the reasoning in the previous part, we get:

$$
\begin{gathered}
K_{1}=0, K_{2}=K, K \leq 64 \\
K_{1}=K-64, K_{2}=64,64<K \leq 104
\end{gathered}
$$

Substituting these values of $K_{1}$ and $K_{2}$ into $M=M_{1}+M_{2}=160-4 K_{1}+16-0.25 K_{2}$.

Joint PPF of two males and one female:

$$
\begin{gathered}
M=176-0.25 K, 0<K \leq 64 \\
M=416-4 K, 64<K \leq 104
\end{gathered}
$$

Answer:

$$
\begin{gathered}
M=176-0.25 K, 0<K \leq 64 \text { (4 points) } \\
M=416-4 K, 64<K \leq 104+\text { graph (4 points) }
\end{gathered}
$$

Arithmetic errors that did not lead to a distortion of the meaning of the problem - penalty 2 points.

Answers without solutions are not credited.