# Solving weekday and weekend fare movies

Thinking if it is a Knapsack problem.

Here is the scenario:

Michael and his daughter (7 year old kid) enjoy going to cinema for movies.

Weekdays fare for kids is 5 dollars, adult 8.
Weekend fare $12.00, regardless kid or adult.  With$100 budget, how can Michael and his daughter enjoy more movies, considering most of those falls in weekdays?

So it costs \$13 for the two of them on a weekday and \$24 for the two of them on a weekend. If you want to maximize the number of movies, skip the expensive ones and the \\$100 budget yields $$\lfloor 100/13 \rfloor = 7$$ movies.

If you prefer writing it explicitly as an optimization (yes, knapsack) problem, let integer decision variables $$x$$ and $$y$$ be the numbers of weekday and weekend movies attended, respectively. The problem is to maximize $$x+y$$ subject to linear constraints \begin{align} 13x+24y &\le 100 \\ x &\ge 0 \\ y &\ge 0 \end{align} The unique optimal solution is $$(x,y)=(7,0)$$, with objective value $$7+0=7$$.

• thank you for the guidance! that's fantastic! Dec 31, 2020 at 6:22