A furninture dealer deals in onlt tables and chairs. He can invest upto Rs 50,000 only has a storage of 100 pieces. Cost price of a table is Rs 1200 and of a chair is Rs 500. He can earn a profit of Rs 180 on the sale of the table and Rs 75 on the sale of one chair. Assuming thatt he can sell all the items he buys, formulate a LP problem so that he can maximize the profit.
Answers
Let,
x be the number of tables.
y be the number of chairs.
As per the question,
The cost of x tables = Rs 1200 x
The cost of y chairs = Rs 500 y
So total cost = 1200 x+500 y.
Also given ,
The storage capacity = 100 pieces.
Thus , x+y ≤ 100
Total investment = Rs 50,000
thus, 1200 x+150 y ≤ 50,000
As the number of chairs can't be negative
Now ,
profit on x tables = Rs 180
profit of y tables = Rs 75
Objective : to maximise the profit..
The Linear programming problem :
Maximise,
Z = 180 x + 75 y
Constraints :
1200 x + 1500 y ≤ 50,000
x + y ≤ 100
x≥0
y≥0
Step-by-step explanation :
★ Consider the -
The quantity of the tables as - 'a'
The quantity of the Charis as - 'b'
Given :
The cost of 'a' tables = Rs 1200 x
The cost of 'b' chairs = Rs 500 y
Total cost = 1200a + 500b.
The storage capacity = 100 pieces.
Total investment = Rs 50,000.
______________________
Now,
The storage capacity = 100 pieces.
It implies that - a + b ≤ 100.
_____________________
Total investment = Rs 50,000.
It implies - 1200a + 150b ≤ 50,000.
______________________
We will find the profit :
Profit on a tables = Rs 180
Profit of b tables = Rs 75
We know that :
★ The Linear programming problem :
The linear programming problem contain a linear function to be minimized or maximized items to certain constraints in the form of linear equations.
__________________
So,
Maximise :
=> 180a + 75b
___________________
Constraints :
=> 1200a + 1500 b ≤ 50,000
=> x + y ≤ 100
=> x ≥ 0
=> y ≥ 0