Created by cian.buckley+1
almost 7 years ago


Copied by PatrickNoonan
almost 7 years ago


A stadium can hold 25 000 people. People attending a regular event at the stadium must purchase a ticket in advance. When the ticket price is €20, the expected attendance at an event is 12 000 people.
(a) If the ticket price was €18, how many people would be expected to attend?
(b) Let x be the ticket price, where x ≤ 20. Write down, in terms of x, the expected attendance at such an event.
(c) Write down a function f that gives the expected income from the sale of tickets for such an event.
(d) Find the price at which tickets should be sold to give the maximum expected income.
(e) Find this maximum expected income.
(f) Suppose that tickets are instead priced at a value that is expected to give a full attendance at the stadium.
(g) The stadium was full for a recent special event.
The results of a survey carried out by the owners suggest that for every €1 reduction, from €20, in the ticket price, the expected attendance would increase by 1000 people.
12000 + (20−18)1000 = 14000
\[12000+(20− x)1000= 32000−1000x\]
\[f(x) = (32000−1000x)x\]
\[f(x) = (32000−1000x)x\]
\[f(x) = 32000−2000x = 0\]
\[x = €16\]
\[f(x) = (32000−1000x)x\]
\[f(16) = (32000−16000)16 = €256000\]
Find the difference between the income from the sale of tickets at this price and the maximum income calculated at (e) above.
\[32000 −1000x = 25000\]
\[1000x = 7000\]
\[x = 7\]
\[f(x) = (32000−1000x)x\]
\[f(7) =(32000−7000)7 =175000\]
\[f(7) =(32000−7000)7 =175000\]
Difference: €256,000 − €175,000 = €81,000
Two types of tickets were sold, a single ticket for €16 and a family ticket (2 adults and 2 children) for a certain amount. The income from this event was €365000.
If 1000 more family tickets had been sold, the income from the event would have been reduced by €14000. How many family tickets were sold?
If 1000 more family tickets had been sold, the income from the event would have been reduced by €14000. How many family tickets were sold?
x = number of single tickets f = number of family tickets y = cost of family ticket x+4f = 25000
\[16x+fy= 365000\]
\[16(x4000)+(f+1000)y = 351000\]
\[16x  64000+fy+ 1000y = 351000\]
\[16x + fy = 365000\]
\[16x + fy = 365000\]
\[1000y=50000\]
\[y=50\]
\[x+4f = 25000\]
\[16x+ 50f = 365000 16x+ 64f = 400000\]
\[16x+ 50f = 365000 16x+ 64f = 400000\]
\[14f = 35000\]
\[f=2500\]
\[f=2500\]
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