Math Problem: Little Caesars - Page 2

On June 06 2012 15:58 figq wrote:
Alright, I think now the setting is finally fully clear to me (thanks for clarifications!), except for this part:
Does that mean all 12 dough balls are finished being flattened and put on the 2 piles before we start using the machine?
Can we switch around stacks, while still flattening the initial dough balls?

On June 07 2012 01:58 netherh wrote:
I don't think this makes sense.

It takes 10 seconds to make a dough ball. It takes 3 seconds to put through the machine. It takes 5 seconds to put in a pan.

It seems these times are fixed for each dough ball, and fiddling with stacking and putting stuff aside won't change that. So as someone else said, making one dough ball at a time is the optimal solution with the information given.

Either that or hiring one more person, so that one person flattens balls (a great job description), and the other person machines flattened balls (an even better job description) and puts them in the pan. Thus you cut the time taken to 10 seconds from 18.

(I think you have to include how long it takes for you to switch tasks, to actually have a problem here to solve...)


Something to note:

Original piece thickness is 2.5cm, and optimal piece thickness is 2.4cm. 60 seconds in stack with 1 on top means decrease of 1/3 size (0.83cm). So 10 seconds in stack with 1 on top means a decrease of 0.14cm.

Since it takes 10 seconds to make the dough piece, the bottom piece in the stack is already too thin when you add the 3rd piece in a stack of 3. This will obviously continue the more you add. So when you have a stack of 12, every single one apart from the top 1 or 2 will be below the proper size.

I bet you're not meant to stack the things at all... you're just lazy. :p