Answer to Question #120138 in Linear Algebra for Peuyehafo

Question

Prices to a concert cost N$200 for an adult and N$150 for a child, the concert venue can
accommodate at most 240 people. The organizers are giving a N$20 discount to every adult
and N$10 discount for every child but do not want the total discounted amount to exceed
N$3200.
Draw a graph and use it to answer the following questions:
i. How many tickets altogether should the organisers sell in order to maximise their sales
amount?
ii. How many tickets of each type (adult and child) should they sell to get a maximum
sales amount?
iii. What is the maximum sales amount the organisers can make?
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?

Answer 1

For adults, concert cost = N$ 200

for child, concert cost = N$ 150

Let x number of tickets sold to adult and y number of tickets are sold to child.

So equation of the problems will be formed as

Cost $= 200x + 150y$

under the constraints

$x +y \leq 240$ and $20x + 10y \leq 3200$ with $x,y \geq 0$

Graph of the equations are shown in figure. Shaded area is the desired solution of the problem, Red line is equation $x +y \leq 240$

Blue line is $20x + 10y \leq 3200$

Points of the boundary (0,240), (160,0) and (80,160)

Among these three points, at point (80,160) Cost will be maximum

i.e. cost = 200*80 + 150*160 = 40000

(i) So total number of tickets sold will be x + y = 80 + 160 = 240

(ii) Adult tickets will be 80 and child tickets will be 160 to maximize the cost or sales amount.

(iii) Maximum sales amount is N$40000

(iv) Total Profit = 40000 - (80*20 + 160*10) = N$36,800