Ski Party

The Situation

The graduating class his year has decided to have a unique graduation party. They have rented a ski hill for a day. The rental costs $250 plus $50 per person. The ski hill has a maximum capacity of 100 people. Explain how this situation can be modelled using a linear relation. Write an equation and graph the resulting line. Discuss the possible restrictions on the variables of this relationship.

Slope

The constant rate of change in this scenario is the $50 per person.

Therefore the slope of the linear model is m = 50

y-intercept

The initial cost, which represents the y-intercept of the linear model, is the $250 rental fee.

Therefore, b = 250

Equation of the line

The equation which describes this relationship is:

y = 50x + 250

where x represents the number of graduates attending the ski party and y represents the total cost of the ski party.

Graph the LineIn your notebook, graph the line on a grid that has a suitable scale.

Do this before proceeding.

Graph the Line

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The Graph

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y = 50x + 250

Restrictions?

In your notebook, write down any restrictions that must exist on this graph.

Revisit the initial problem to see if there is any information there that could restrict our graph.

Do this before proceeding.

Restrictions

There are several restrictions on this situation.

Firstly, since it is impossible to have a part of a graduate attend the party, the x variable must be a positive integer. (No negative values of x are allowed.)

Restrictions

Secondly, since the ski hill has a maximum capacity of 100 people, the x variable can be no higher than 100.

Therefore, . (We say, “x is greater than or equal to zero and less than or equal to 100”)

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New GraphOur final graph should show the restrictions.

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y = 50x + 250

Maximum cost

Now, getting back to the maximum cost issue:

A maximum of 100 people creates a maximum cost for the ski party:

y = 75(100) + 250

y = 7500 + 250

y = 7750

Therefore the maximum cost of the total cost of the ski party is $7750.