Math, asked by Rythm14, 11 months ago

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

Answered by Anonymous
42

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


Rythm14: tq gunie :)
Answered by Blaezii
42

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


Rythm14: tq :P
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