What helps drivers retain steering control during emergency braking?
2 answers:
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Answer:
i. 240 tickets
ii. 80 adults and 160 children
iii. N$ 40,000.00
iv. Profit= N$ 36,800.00
Explanation:
Ok so in order to solve this problem we need to start by building our system of equations from the restrictions the problem gives us:
let's say that
x=# of adult tickets
y= # of children tickets
so
"... the concert venue can accommodate at most 240 people."
this translates to the following inequality:
"The organizeres are giving a N$20 discount to every adult and N$10 discount for every child but do not want the total descounted amount to exceed N$3200."
Translates to
and we know we cannot sell a negative number of adult and children tickets so the last restrictions are:
so we can now graph our system of inequalities (see graph attached)
the procedure to graph a linear equation is as follows:
1. Pick an x-value
2. Find the corresponding y-value with the provided equation we need to graph.
3. Write the coordinates as an ordered pair (x,y)
4. Repeat the process with a second x-value.
5. Graph the two points you found on the previous steps.
6. Connect the points with a straight line.
So the feasible region is the one that all restrictions will have in common and the maximum and minimum points for our objective function will be the marked points, which are the vertices of the feasible region. So we test them to see which will maximize our objective function.
Objective function:
s=200x+150y
(0,0)
s=200(0)+150(0)=0
(0,240)
s=200(0)+150(240)=N$36,000
(80,160)
s=200(80)+150(160)=N$40,000
(160,0)
s=200(160)+150(0)=N$32,000
So when comparing the results we can see that the objective function will be maximized at the point (80,160) so:
i) How many tickets altogether should the organisers sell in order to maximize their sales amount?
total # of tickets = x+y = 80+160=240 tickets
ii) How many tickets of each type (adult and child) should they sell to get a maximum sales amount?
x y
80 adult tickets and 160 child tickets.
iii) What is the maximum sales amount the organizers can make?
s=200(80)+150(160)=N$40,000
iv) If the profit is calculated as follows: profit = total sales - total discount amount. How much profit will the organisers make from the maximum sales?
Profit= total sales - total discount
total sales=200(80)+150(160)=N$40,000
total discount = 20(80)+10(160)=N$3,200
Profit=N$40,000-N$3,200=N$36,800.00
