Question

a shopkeeper deals in two items-thermos flasks and air tight containers. a flask costs him $rs.$ 120 and an air tight container costs him $rs.$ 60. he has at the most $rs.$ 12000 to invest and has space to store a maximum of 150 items. the profit on selling a flask is $rs.$ 20 and an air tight container is $rs.$ 15. assuming that he will be able to sell all things he buys, how many flasks and containers respectively should he buy to maximize his profit?

Answer 1

Answer:

The shopkeeper need to by at least 50 flasks 100 airtight containers  to suit his rs 12000 budget.

Step-by-step explanation:

For us to solve the problem ,  

we let x represent  the thermos flasks

then Y will represent the air-tight containers,

We now form  a set of simultaneous equations to represent the information provided in the problem

12000=120x+60y…………………….(i)

150=x+y………………………………….(ii)

Th first equation represents the value ( Rs 12000) of buying the items at the cost of Rs 120 for a thermos and Rs 60 for an airtight container

The second equation represents the total number of flasks and containers (150) the shopkeeper intends to buy.

Let us now solve the simultaneous equations.

Using equation  (ii) above , we can find the value of x. that is  

150= x+y

150-y=x………(iii)

From equation (iii), we substitute the value of x to equation(i)

Therefor we obtain ;

12000 = 120(150-y) + 60 y

12000 = 18000-120y + 60y

18000-12000 =60y

6000 = 60y

100=y  

Thus the value of y becomes 100, implying that the shopkeeper needs to buy 100 airtight containers for the 12000 to fit his budget

Now lets find the value of x,

Substituting the value of y in equation (iii) to obtain x

150-y =x

150-100=x

X=50

Implying that the shopkeeper need to by at least 50 flasks to suit his budget.