Practicing ML Aggarwal Class 12 Solutions Chapter 3 Linear Programming Ex 3.1 is the ultimate need for students who intend to score good marks in examinations.

## ML Aggarwal Class 12 Maths Solutions Section C Chapter 3 Linear Programming Ex 3.1

Question 1.
Two tailors, A and B, earn f 300 and ₹ 400 per day respectively. A can stitch 6 shirts and 4 pairs of trousers while B can stitch 10 shirts and 4 pairs of trousers per day. To find how many days should each of them work and if it is desired to produce atleast 60 shirts and 32 pairs of trousers at a minimum cost labour, formulate this as an LPP.
From the given data we make the following table

 Tailors items per day A B Minimum requirement Shirts 6 10 60 pair of trousers 4 4 32

given earning of tailors A and B per day be ₹ 300 and ₹ 400y.
Thus earningof tailor A in x days = ₹ 300x
and earning of Tailor B in y days = ₹ 400y
Let Z be the table labour cost which is to be minimized
Min Z = 300x + 400y
Thus shirt constraint be 6x + 10y ≥ 60 i.e. 3x + 5y ≥ 30
Trouser’s constraint be 4x + 4y ≥ 32 i.e. x + y ≥ 8
and non-negativity constraints x ≥ 0, y ≥ 0 [since days can’t be negative]
Thus the mathematical modelling of given LPP be as under min Z = 300x + 400y
Subject to constraints ;3x + 5y ≥ 30; x + y ≥ 8; x ≥ 0 ; y ≥ 0

Question 2.
A small firm manufacturers necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is atmost 24. It takes one hour to make a bracelet and half an hour to make a necklace. The maximum number of hours available per day is 16. If the profit on a necklace is ₹ 100 and that on a bracelet is ₹ 300, formulate an L.P.P. for finding how many of each should be produced daily to maximize the profit ? It being given that atleast one of each must be produced.
Let x be the number of necklaces and y be the number of brackets manufactured by a small firm. Since objective of the manufacturer to maximise the profit. Since profit gained by manufacturer on selling on one necklace and one bracelet is given to be ₹ 100 and ₹ 300. Thus profit gained by manufacturer on x necklaces and y bracelets be ₹ 100x and ₹ 300y.
Thus total profit of manufacturer = Z = 100x + 300y
We have to maximise Z.
Given total no.of necklaces and bracelets made by manufacturer per day be atmost 24. i.e. x + y < 24
Time taken by manufacturer to make one bracelet and one necklace be one hour and half an hour respectively. Thus time take by manufacturer to make y bracelets and x necklaces be y hours and $$\frac{x}{2}$$ hours and maximum time available per day be 16 hours x.
∴ $$\frac{x}{2}$$ + y ≤ 16
⇒ x + 2v ≤ 32
Also necklaces and bracelets cannot be negative and also it is given that, atleast one necklace and one bracelet must be produced by manufacturer, i.e. x ≥ 1, y ≥ 1 Thus, the mathematical modelling of given LPP is as under
Max Z = 100x + 300y Subject to constraints :
⇒ x + y ≤ 24
⇒ x + 2y ≤ 32
⇒ x ≥ 0, y ≥ 0

Question 3.
A furniture dealer deals in only two items — tables and chairs. He has ₹ 20000 to invest and a space to store atmost 80 pieces. A table costs him ₹ 800 and a chair costs ₹ 200. He can sell a table for ₹ 950 and a chair for ₹ 280. Assume that he can sell all the items that he buys. Formulate this problem as an L.P.P. so that he can maximize his profit.
Let x be the number of chairs andy be the number of tables that a furniture dealer buys and sells.
Now cost of each table = ₹ 800
S.P of each table = ₹ 950
∴ profit of dealer on each table = ₹ (950 – 800) = ₹ 150
again cost of each chair = ₹ 200
S.P of each chair = ₹ 280
profit of dealer on each chair = ₹ (280 – 200) = ₹ 80
Thus, the total profit of the dealer on selling x tables and y chairs = 150x + 80y
Let Z be the total profit of dealer
Then Z = 150x + 80y which is the objective function
So cost of x tables and y chairs be 800x + 200y and the dealer can invest atmost ₹ 20000.
Thus, the investment constraint is 800x + 200y < 20000 i.e. 4x + y ≤ 100
Also the dealer has a space to store atmost 80 pieces
∴ x + y ≤ 80 (space constraint)
Thus, the mathematical formulation of LPP is given below :
Max Z = 150x + 80y Subject to constraints :
4x +y ≤ 100 ;
x + y ≤ 80
x, y ≥ 0
[Since the number of chairs and tables cannot be negative]