Linear Programming Exercise 02
Question:
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14kg of nitrogen and 14kg of phosphoric acid for her crop. If F1 cost Rs.6/kg and F2 costs Rs.5/kg, determine how much of each type of fertilizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost?
Answer:
Step 1: Calculate the amount of F1 and F2 required to meet the nutrient requirements:
F1 required = (14kg of nitrogen/10%) + (14kg of phosphoric acid/6%) = 16.8kg + 2.33kg = 19.13kg
F2 required = (14kg of nitrogen/5%) + (14kg of phosphoric acid/10%) = 28kg + 1.4kg = 29.4kg
Step 2: Calculate the minimum cost:
Minimum cost = (19.13kg × Rs.6/kg) + (29.4kg × Rs.5/kg) = Rs.114.78 + Rs.147 = Rs.261.78
Question:
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A required 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B required 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours of assembling. The profit is Rs.5 each for type A and $Rs.6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit?
Answer:
 Calculate the number of souvenirs that can be manufactured in 3 hours 20 minutes of cutting and 4 hours of assembling:
Type A: Cutting: (3 hours 20 minutes) ÷ (5 minutes per souvenir) = 64 souvenirs Assembling: (4 hours) ÷ (10 minutes per souvenir) = 24 souvenirs
Type B: Cutting: (3 hours 20 minutes) ÷ (8 minutes per souvenir) = 40 souvenirs Assembling: (4 hours) ÷ (8 minutes per souvenir) = 30 souvenirs

Calculate the total profit for each type of souvenir: Type A: 64 souvenirs x Rs.5 = Rs.320 Type B: 40 souvenirs x Rs.6 = Rs.240

Calculate the total profit for both types of souvenir: Rs.320 + Rs.240 = Rs.560

Determine the number of souvenirs of each type that will maximise the profit: Type A: 64 souvenirs Type B: 40 souvenirs
Question:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.(i) What number of rackets and bats must be made if the factory is to work at full capacity?
Answer:
Step 1: Calculate the total number of hours required to make one tennis racket and one cricket bat.
Total hours required for one tennis racket = 1.5 hours of machine time + 3 hours of craftsman’s time = 4.5 hours
Total hours required for one cricket bat = 3 hours of machine time + 1 hour of craftsman’s time = 4 hours
Step 2: Calculate the maximum number of tennis rackets and cricket bats that can be made in a day.
Maximum number of tennis rackets that can be made in a day = 42 hours of machine time/1.5 hours of machine time per racket = 28 rackets
Maximum number of cricket bats that can be made in a day = 24 hours of craftsman’s time/1 hour of craftsman’s time per bat = 24 bats
Step 3: Calculate the total number of rackets and bats that can be made in a day.
Total number of rackets and bats that can be made in a day = 28 rackets + 24 bats = 52 rackets and bats
Question:
A factory manufactures two types of screws, A and B. Each type screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs.7 and screws B at a profit of Rs.10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise profit? Determine the maximum profit.
Answer:
Step 1: Calculate the number of packages of each type of screw that can be produced in a day using the available machines.
Number of packages of screws A = (4 min × 2 machines × 4 hours)/4 min = 32 packages
Number of packages of screws B = (6 min × 2 machines × 4 hours)/3 min = 48 packages
Step 2: Calculate the total profit from selling each type of screw.
Profit from selling screws A = 32 packages × Rs.7 = Rs.224
Profit from selling screws B = 48 packages × Rs.10 = Rs.480
Step 3: Calculate the maximum profit by adding the profits from selling screws A and B.
Maximum profit = Rs.224 + Rs.480 = Rs.704
Question:
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at least for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs.5 and that from a shade is Rs. 3 . Assuming that the manufacturer can sell all the lamps and shades that the produces , how should he schedule his daily production in order to maximise his profit?
Answer:
Step 1: Calculate the total time taken to produce each item.
To produce a pedestal lamp, it takes 2 hours on the grinding/cutting machine and 3 hours on the sprayer, so the total time taken to produce a lamp is 5 hours.
To produce a shade, it takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer, so the total time taken to produce a shade is 3 hours.
Step 2: Calculate the maximum number of items that can be produced in a day.
The sprayer is available for at least for at the most 20 hours and the grinding/cutting machine for at the most 12 hours.
Therefore, the maximum number of lamps that can be produced in a day is (20 hours/5 hours) = 4 lamps.
The maximum number of shades that can be produced in a day is (12 hours/3 hours) = 4 shades.
Step 3: Calculate the maximum profit that can be earned in a day.
The profit from the sale of a lamp is Rs.5 and that from a shade is Rs. 3.
Therefore, the maximum profit that can be earned in a day is (4 lamps x Rs.5) + (4 shades x Rs.3) = Rs.32.
Step 4: Schedule the daily production in order to maximise the profit.
The manufacturer should produce 4 lamps and 4 shades in a day in order to maximise his profit.
Question:
One kind of cake required 200 g flour and 25 g of fat, and another kind of cake required 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes?
Answer:
Step 1: Determine the amount of flour and fat available.
5 kg of flour and 1 kg of fat are available.
Step 2: Convert the available flour and fat into grams.
5 kg of flour = 5,000 g of flour
1 kg of fat = 1,000 g of fat
Step 3: Calculate the amount of flour and fat required for each cake.
One kind of cake requires 200 g of flour and 25 g of fat.
Another kind of cake requires 100 g of flour and 50 g of fat.
Step 4: Calculate the maximum number of cakes that can be made from the given amount of flour and fat.
For the first kind of cake:
5,000 g of flour ÷ 200 g of flour = 25 cakes
1,000 g of fat ÷ 25 g of fat = 40 cakes
For the second kind of cake:
5,000 g of flour ÷ 100 g of flour = 50 cakes
1,000 g of fat ÷ 50 g of fat = 20 cakes
Step 5: Determine the maximum number of cakes that can be made.
The maximum number of cakes that can be made is 25, as this is the lowest number of cakes that can be made with the available flour and fat.
Question:
A merchant plans to sell two types of personal computers, a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant would stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
Answer:

The merchant’s total budget is Rs 70 lakhs.

The maximum monthly demand of computers is 250 units.

The cost of desktop model is Rs 25000 and the cost of portable model is Rs 40000.

The profit on the desktop model is Rs 4500 and the profit on the portable model is Rs 5000.

The merchant should stock the maximum number of desktop models that he can afford without exceeding his budget.

The merchant can stock 250 desktop models at a cost of Rs 25 lakhs (25000 x 250 = 25 lakhs).

The remaining budget of Rs 45 lakhs can be used to buy portable models.

The merchant can buy 1125 portable models at a cost of Rs 45 lakhs (40000 x 1125 = 45 lakhs).

Therefore, the merchant should stock 250 desktop models and 1125 portable models to get maximum profit.
Question:
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit, of Rs.17.50 per package on nuts and Rs.7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day?
Answer:
Step 1: Let x = number of packages of nuts and y = number of packages of bolts produced each day.
Step 2: Write the objective function.
The objective is to maximize the profit, which is given by: Profit = 17.50x + 7.00y
Step 3: Write the constraints.
The constraints are given by:
 Time on machine A: 1x + 3y ≤ 12
 Time on machine B: 3x + 1y ≤ 12
 Nonnegativity: x ≥ 0, y ≥ 0
Step 4: Solve the problem using the graphical method.
The graphical method can be used to solve this problem. The solution is x = 4 and y = 4, so the manufacturer should produce 4 packages of nuts and 4 packages of bolts each day to maximize his profit.
Question:
The corner points of the feasible region determined by the following system linear inequalities: 2x+y≤10,x+3y≤15,xy≥0 are (0,0),(5,0),(3,4) and (0,5). Let Z=px+qy, where p,q>0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is A p=q B p=2q C p=3q D q=3p
Answer:
A p=q
Question:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftsman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time. If the profit on a racket and on a bat is Rs.20 and Rs.10 respectively, find the maximum profit of the factory when it works at full capacity
Answer:

Let x represent the number of tennis rackets and y represent the number of cricket bats.

The total machine time and craftsman’s time used should not exceed the availability of 42 hours and 24 hours respectively.

Therefore, the constraint equations are: 1.5x + 3y ≤ 42 3x + y ≤ 24

The objective function is: Maximize Z = 20x + 10y

The solution of the above linear programming problem is: x = 14, y = 8

Therefore, the maximum profit of the factory when it works at full capacity is Rs.280 (20x + 10y = 2014 + 108 = 280).
Question:
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs Rs.4 per unit food and F2s costs Rs.6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the mineral nutritional requirements?
Answer:
Let x = number of units of food F1 Let y = number of units of food F2
Objective Function: Minimize Cost = 4x + 6y
Constraints: 3x + 6y ≥ 80 4x + 3y ≥ 100 x, y ≥ 0
Linear Programming Problem: Minimize Cost = 4x + 6y Subject to: 3x + 6y ≥ 80 4x + 3y ≥ 100 x, y ≥ 0
Question:
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs. 60/kg and Food Q costs Rs. 80/kg. Food P contains 3 units/kg of vitamin A and 5 units/kg of vitamin B while food Q contains 4 units/kg of vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Answer:
Step 1: Find the total number of units of vitamin A and B in the mixture.
Total units of Vitamin A = (3 units/kg of Food P) + (4 units/kg of Food Q) Total units of Vitamin B = (5 units/kg of Food P) + (2 units/kg of Food Q)
Step 2: Calculate the minimum amount of Food P and Food Q required to meet the minimum vitamin content requirement.
Minimum amount of Food P required = (8 units of Vitamin A)/(3 units/kg of Food P) = 2.67 kg Minimum amount of Food Q required = (11 units of Vitamin B)/(2 units/kg of Food Q) = 5.5 kg
Step 3: Calculate the minimum cost of the mixture.
Minimum cost of the mixture = (2.67 kg of Food P x Rs. 60/kg) + (5.5 kg of Food Q x Rs. 80/kg) = Rs. 476