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Maths problem 2.

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Presentation on theme: "Maths problem 2."— Presentation transcript:

1 Maths problem 2

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?

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

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

5 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.

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

7 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.


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