Canadas Wonderland Admission Prices
Linear systems Assignment
- Buy a season's pass without a parking pass
- Buy a season's pass with a parking pass
- Don't buy a season's pass or a parking pass
He will probably go to Wonderland once a week during his summer vacation. This means that he will be visiting Canada's Wonderland 8 times over the summer. Which option should he go with, 1, 2 or 3?
What we Know
Parking Pass $40
Season's Pass $70
One-time Admission Fee $40/visit
One-time Parking Fee $20/visit
Situation 1 ► Season's Pass without a Parking Pass
C = Season's Pass + Parking Fee / visit
∴ y = 20x + 70
Situation 2 ► Season's Pass with a Parking Pass
C = Season's Pass + Parking Pass
∴ y = 70 + 40
∴ y = 110
Situation 3 ► No Season's Pass or Parking Pass
C = Admission Fee / visit + Parking Fee / visit
C = $40 / visit + $20 / visit
C = ($40 + $20) / visit
C = $60 / visit
∴ y = 60x
The y-axis represents ► Money Spent ($)
The x-axis represents ► Number of Visits
When I made the graph i divided all the equations by 10 to be able to represent the situation in an enhanced manner (visually). This way it is easier to pinpoint the locations at which these lines intercept.
Situation 1 is represented by the RED line
Situation 2 is represented by the BLUE line
Situation 3 is represented by the GREEN line
Lines 1 and 2 intersect at (2,110).
Lines 1 and 3 intersect at (1.75,105)
Lines 2 and 3 intersect at (1.8,110)
The coordinates at which these lines meet mark the point at which it would cost John the same amount of money (x) as long as he took the specific number of visits (y) whether he took any of those 2 situations.
What's the best Situation for John?
Prices for every situation if John visits Wonderland 8 times:
Situation 1 ►
y = 20(8) + 70
y = 160 + 70
y = 230
Situation 2 ►
y = 110
Situation 3 ►
y = 60(8)
y = 480
We can clearly see that Situation 2 is the cheapest option for John.
John should buy both, a Season's Pass and a Parking Pass