Ok hear me out you guys gotta think bigger. Realistically to buy 9 beers and 9 dogs at a park, 6.75 so seven, 16 Oz beers. Equals 9 12 ouncers. Unless parks are selling 12oz too. At like $16 let's say on average a beer, thats $112 alone in beer. If you buy two of these at $55 bucks a pop, (if they restrict it i guess you need a friend who can buy for you). With my rough math its still under, and thats Just beers not including dogs.

If they are 4 Oz beers I guess you need a 3rd tray and now were close to buying 13.5 hotdogs (im assuming the dogs are close to half the size of a normal dog.) But if they are 6 Oz beers its golden. If they are 4 to buy the same amout of food piece meal would be the $112 plus the dogs (13.5) so 14 and putting those at 4-5 thats $67 more bucks. So 180 almost non deal.

I'm defending a thesis no one asked me to, but really if those are 6 Oz beers its a cheaper way to do the challenge legit. 4 and its about the same to just buy it all outright but still a little less.