Respuesta :
Let x be the number of toy truck, y is the number of toy car, and z is the number of toy boat.
The constraint for the time
[tex]10x+12y+8z\le 400[/tex]
The constraint for the profit
[tex]x+1.50y+0.6z\ge 35[/tex]
The constraint for the time
[tex]10x+12y+8z\le 400[/tex]
The constraint for the profit
[tex]x+1.50y+0.6z\ge 35[/tex]
Answer:
[tex]\left\{\begin{array}{l}10x+12y+8z\le 400\\x+1.5y+0.6z\ge 35\end{array}\right.[/tex]
Step-by-step explanation:
Let x be the number of toy trucks, y be the number of toy cars and z be the number of toy boats a worker makes each day.
1. If a worker spends 10 minutes to make one toy truck, then he spends 10x minutes to make x toy trucks. If a worker spends 12 minutes to make one toy car, then he spends 12y minutes to make y toy cars. If a worker spends 8 minutes to make one toy boat, then he spends 8z minutes to make z toy boats. In total he can spend at most 400 minutes each day, then
[tex]10x+12y+8z\le 400.[/tex]
2. If a worker generates profit of $1.00 per one toy truck, then he generates profit of $x per x toy trucks. If a worker generates profit of $1.50 per one toy car, then he generates profit of $1.50y per y toy cars. If a worker generates profit of $0.60 per one toy boat, then he generates profit of $0.60z per z toy boats. The manufacturer needs each worker to generate a potential profit of $35 each day, then
[tex]x+1.5y+0.6z\ge 35.[/tex]
Thus, the system of two inequalities is
[tex]\left\{\begin{array}{l}10x+12y+8z\le 400\\x+1.5y+0.6z\ge 35\end{array}\right.[/tex]