A factory that produces tables and chairs is equipped with 10 saws, 6 lathes, and 18 sanding machines. It takes a chair 10 minutes on a saw, 5 minutes on a lathe, and 5 minutes of sanding to be completed. It takes a table 5 minutes on a saw, 5 minutes on a lathe, and 20 minutes of sanding to be completed. A chair sells for 10 and a table sells for 20. How many tables and chairs should the factory produce per hour to yield the highest revenue, and what is that revenue?
Solution
总共有 \(60min\), 由于上述条件是对于一台机器而言的，所以对于多个机器，可利用的时间即为 \(n\cdot 60 min\).
设 \(chair = x, table = y\), 我们需：
\[\max 10x+20y \]限制条件为：
\[\begin{align} 10x+5y&\le 60*10\\ 5x+5y&\le 60*6\\ 5x+20y&\le 60*18 \end{align} \]候选答案就在这些线的交点处，所以求解每个交点，然后取最大的 \(10x+20y\) 即可
标签：10,20y,28,MathProblem,factory,sanding,60,problem,minutes From： https://www.cnblogs.com/xinyu04/p/16584713.html