3. Two females with PPF $M=40-2K$. PPFs are linear with identical opportunity costs. By adding individual PPFs, we get that the PPF of the two females: M=80-2K, $\mathrm{K} \leq 40$.
Answer: $\mathrm{M}=80-2 \mathrm{~K}, \mathrm{~K} \leq 40$. (1 point)

Two males with the production possibility frontier (PPF) M=64-K^2. The PPFs have monotonically increasing alternative costs, which coincide at each K_1=K_2. To obtain the maximum amount of M for each value of K, choose K_1=K_2=0.5K. The joint PPF of the two males will be:

$\mathrm{M}=64-\mathrm{K} \_1 \wedge 2+64-\mathrm{K}_{-} 2 \wedge 2=128-0.25 \mathrm{~K}^{\wedge} 2-0.25 \mathrm{~K} \wedge 2=128-0.5 \mathrm{~K}^{\wedge} 2$

Answer: $\mathrm{M}=128-0.5 \mathrm{~K}^{\wedge} 2, \mathrm{~K} \leq 16$ (3 points)

A male with the PPF $М=64-\mathrm{K}^{\wedge} 2$ and a female with the PPF $\mathrm{M}=40-2 \mathrm{~K}$. The alternative costs of the male are AI=2K_1, and the female's AI=2. We get that

$\mathrm{K} \leq 1: \mathrm{K} \_1=\mathrm{K}, \mathrm{K} \_2=0 ;$

$1<\mathrm{K} \leq 21: \mathrm{K}_{-} 1=1, \mathrm{~K} \_2=\mathrm{K}-1$

$21<\mathrm{K} \leq 28: K \_1=\mathrm{K}-20, \mathrm{~K} \_2=20$.

Substituting these values into the function M=M_1+M_2=64-К_1^2+40-2К_2, we get the joint PPF of the male and female:

$\mathrm{M}=104-\mathrm{K}^{\wedge} 2, \mathrm{~K} \leq 1$

$\mathrm{M}=105-2 \mathrm{~K}, 1<\mathrm{K} \leq 21$

$\mathrm{M}=40 \mathrm{~K}-\mathrm{K} \wedge 2-336,21<\mathrm{K} \leq 28$

Answer: $\mathrm{M}=104-\mathrm{K}^{\wedge} 2, \mathrm{~K} \leq 1$

$\mathrm{M}=105-2 \mathrm{~K}, 1<\mathrm{K} \leq 21$

$\mathrm{M}=40 \mathrm{~K}-\mathrm{K} \wedge 2-336,21<\mathrm{K} \leq 28$ (5 points)