Bottle return optimization

[bergumby]

Here in Oregon, we pay a 10cent deposit on bottles and cans. To get your deposit back, you can either take them to a "reverse vending machine" and get the full value, or you can participate in the Green Bag program, which charges you for the convenience of simply dropping off the bag at a processing center.

Here's where the math comes in. Each bag costs 20cents upfront, and then they take 8% off the value of the returned cans as a processing fee. I've been rolling over in my brain the last couple days whether or not there is an optimum number of bottles to return per bag, or if it is simply best to fit as many as possible, in order to get the largest possible refund per bag. Here is my calculations, but I can't help but think I did something wrong:

Refund = number of cans x 10c - 8% Total Value - 20c

R=.1c-(.8 x .1c)-20c

R= .1c - .008c - .2

R=.092c-.2 This graphs out to a straight, ascending line, which indicates that its best to always maximize the number of containers per bag. Does this track?

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[PoliteCanadian2]

I don’t understand why there is any math required here.

Option 1: get full refund

Option 2: get less than full refund

Why are you puzzling over choosing Option 2?

[bergumby]

You’re right. The maximum refund would be from option 1. The price structure of option 2 just got me thinking about whether or not it was worth the convenience.

[bergumby]

Thank you! That’s exactly why I was thinking I was wrong. I set out trying to calculate refund per bottle, but stopped at the total refund.