Answer:
a) Let x represent the the number of tickets given out for the seats that cost $7 and let y represent the the number of tickets given out for the seats that cost $10
b) x + y ≥ 8...(1)
7×x + 10×y ≤ 70...(2)
c) Please see the attached graph
d) One possible solution is which is within the feasible region is x = 4 and y = 4.
Step-by-step explanation:
The given parameters are;
The prices of the tickets for the seats given by the radio station are $7 and $10
The minimum number of tickets to be given out = 8 tickets
The maximum number of tickets the radio station can give away = $70
a) Let x represent the the number of tickets given out for the seats that cost $7 and let y represent the the number of tickets given out for the seats that cost $10
b) x + y ≥ 8...(1)
7×x + 10×y ≤ 70...(2)
The above inequalities can be written with respect to y as follows;
For the first inequality, we have;
y ≥ 8 - x
For the second inequality, we have;
7×x + 10×y ≤ 70
y ≤ 70/10 - 7/10×x
Which gives;
y ≤ 7 - 7/10×x
c) The two inequalities are graphed as using Microsoft Excel as follows;
With the given inequalities expressed in terms of y,
1) A table of values is generated for each inequality expressed as equations
2) The table is plotted using the insert function of Microsoft Excel
3) The area where the valid region of both inequalities intersect is shaded
d) The possible solution is given by the point where both inequality equations are equal as follows;
8 - x = 7 - 7/10×x
8 - 7 = x - 7/10×x
3/10×x = 1
x = 10/3 = 3¹/₃
y = 7 - 7/10×x = 7 - 7/10×10/3 = 14/3 = 4²/₃
Therefore, one possible solution is which is within the feasible region is x = 4 and y = 4.
