Maths problem 2

Charge – problem: best dog sitter A dog sitting company offer two deals: Deal 1: fix costs for 100€ and 20€ per hour to look after your dogs in the holidays. Deal 2: fix costs for 200€ and 5€ per hour Which is the better deal?

X – time for hiring the company Deal 1: 100+20x Deal 2: 200+5x F1(x)= 100+20x F2(x)= 200+5x

The solution F1(x)=F2(x) 100+20x=200+5x 20x-5x=200-100 15x=100 /:15 X=100:15=6,67h=6h40m (0,67h=0,67.60=40,2min)

Conclusion Until 6h 40 min the first offer is better, 1) If x<6h 40min then F1(x)<F2(x) If the time is less than 6h 40min the first offer is better. 2) If x=6h 40min then F1(x)=F2(x) When the time is 6h 40min both offers are equally expensive. 3) If x>6h 40min then F1(x)>F2(x) If the time is more than 6h 40min the second offer is better. Until 6h 40 min the first offer is better, after 6h 40 min the second offer is better.

THANK YOU FOR THE ATTENTION Created by: Ventsislav Atanasov, Velislava Stoycheva, Valeria Hristova and Elmira Karadacheva. 

This project has been funded with support from the European Commission. This publication [communication] reflects the views only of the author, and the Commission cannot be held responsible for any use which may be made of the information contained therein.