CBSE Class 12 Mathematics Linear Programming Worksheet Set 01

CBSE Class 12 Mathematics Linear Programming (1). CBSE issues sample papers every year for students for class 12 board exams. Students should solve the CBSE issued sample papers to understand the pattern of the question paper which will come in class 12 board exams this year. The sample papers have been provided with marking scheme. It’s always recommended to practice as many CBSE sample papers as possible before the board examinations. Sample papers should be always practiced in examination condition at home or school and the student should show the answers to teachers for checking or compare with the answers provided. Students can download the sample papers in pdf format free and score better marks in examinations. Refer to other links too for latest sample papers.

Question. A Linear function, which is minimized or maximized is called
(a) an objective function
(b) an optimal function
(c) A feasible function
(d) None of these
Answer : A

Question. Any feasible solution which maximizes or minimizes the objective function is Called:
(a) A regional feasible solution
(b) An optimal feasible solution
(c) An objective feasible solution
(d) None of these
Answer : B

Question. The maximum value of Z = 3x + 4y subject to the constraints : x+ y ≤ 4, x ≥ 0 , y ≥ 0 is :
(a) 0
(b) 12
(c) 16
(d) 18
Answer : C

Question. The maximum value of 𝑍 = 4𝑥 + 2𝑦 subjected to the Constraints2𝑥 + 3𝑦 ≤ 18 ,𝑥 + 𝑦 ≥ 10 ;𝑥, 𝑦 ≥ 0 is
(a) 320
(b) 300
(c) 230
(d) none of these
Answer : D

Question. The point in the half plane 2𝑥 + 3𝑦 − 12 ≥ 0 is :
(a) (- 7,8 )
(b) ( 7 , - 8 )
(c) ( -7 , - 8 )
(d) (7, 8 )
Answer : D

Question. The solution set of the in equation 2𝑥 + 𝑦 > 5 is
(a) Half plane that contains the origin
(b) Open half plane not containing the origin
(c) Whole 𝑥𝑦 −plane except the points lying on the line 2𝑥 + 𝑦 = 5
(d) None of these
Answer : B

Question. Objective function of a LPP is
(a) a constraint
(b) a function to be optimized
(c) a relation between the variables
(d) none of these
Answer : B

Question. Which of the following statements is correct?
(a) Every L P P admits an optimal solution
(b) A L P P admits unique optimal solution
(c) If a L P P admits two optimal solution solutions, it has aninfinite number of optimal solutions
(d) The set of all feasible solutions of a LPP is a finite set.
Answer : C

Question. The maximum value of Z = 2x +3y subjectto the constraints :
𝑥 + 𝑦 ≤ 1 , 3𝑥 + 𝑦 ≤ 4 , 𝑥, 𝑦 ≥ 0is
(a) 2
(b) 4
(c) 5
(d) 3
Answer : C

Question. The optimal value of the objective function is attained at the points :
(a) Given the intersection of inequations with the axes only
(b) Given by intersection of inequations with X-axis only
(c) Given by corner points of the feasible region
(d) None of these.
Answer : C

CASE STUDY QUESTIONS

I. A company started airlines business and for running business it bought aeroplanes . Now an aeroplane can carry maximum of 200 passengers . A profit of Rs.400 is made on each first class ticket and a profit of Rs.300 is made on each second class ticket . The airline reserves at least 20 seats for first class .However , at least four times as many passengers prefer to travel by second class then by first class . Company wants to make maximum profit by selling tickets of first class (𝑥) and second class (𝑦) .
Using the above information give the answer of the following questions.

Question. To get maximum profit how many first class tickets should be sold –
(a) 20
(b) 180
(c) 160
(d) 40
Answer : D

Question. Corner points of feasible region are
(a) (20,180)
(b) (20,0)
(c) (40,0)
(d) all the above
Answer : D

Question. Difference between the maximum profit and minimum profit is equal to
(a) 8000
(b) 56000
(c) 64000
(d) none of the above
Answer : A

Question. The objective function is
(a) 400𝑥 + 300𝑦
(b) 300𝑥 + 400𝑦
(c) 𝑥 + 𝑦
(d) none of the above
Answer : A

Question. Minimum profit is equal to
(a) 8000
(b) 6000
(c) 64000
(d) none of the above
Answer : A

Short Answer Questions

Question. A small firm manufactures necklaces and bracelets. The total number of necklaces and bracelets that it can handle per day is at most 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 Rs. 100 and that on a bracelet is Rs. 300. Formulate an LPP. for finding how many of each should be produced daily to maximise the profit? It is being given that at least one of each must be produced. 
Answer: Let \( x \) and \( y \) be the number of necklaces and bracelets manufactured by small firm per day. If \( P \) be the profit, then objective function is given by
\( P = 100x + 300y \) which is to be maximised under the constrains
\( x + y \leq 24 \)
\( \frac{1}{2}x + y \leq 16 \)
\( x \geq 1, y \geq 1 \)

Question. Two tailors, A and B, earn Rs. 300 and Rs. 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 at least 60 shirts and 32 pairs of trousers at a minimum labour cost, formulate this as an LPP. 
Answer: Let A and B work for \( x \) and \( y \) days respectively.
Let \( Z \) be the labour cost.
\( Z = 300x + 400y \)
Subject to constraints
\( 6x + 10y \geq 60 \)
\( 4x + 4y \geq 32 \)
\( x, y \geq 0 \)

Question. A company produces two types of goods A and B, that require gold and silver. Each unit of type A requires 3 g of silver and 1 g of gold while that of type B requires 1 g of silver and 2 g of gold. The company can produce a maximum of 9 g of silver and 8 g of gold. If each unit of type A brings a profit of Rs. 40 and that of type B Rs. 50, formulate LPP to maximize profit. 
Answer: Let \( x \) and \( y \) be the number of goods A and goods B respectively. If \( P \) be the profit then
\( P = 40x + 50y \) which is to be maximised under constraints
\( 3x + y \leq 9 \)
\( x + 2y \leq 8 \)
\( x \geq 0, y \geq 0 \)

Question. A firm has to transport atleast 1200 packages daily using large vans which carry 200 packages each and small vans which can take 80 packages each. The cost for engaging each large van is Rs. 400 and each small van is Rs. 200. Not more than Rs. 3,000 is to be spent daily on the job and the number of large vans cannot exceed the number of small vans. Formulate this problem as a LPP given that the objective is to minimize cost. 
Answer: Let the number of large vans and small vans be \( x \) and \( y \) respectively.
Here transportation cost \( Z \) be objective function, then
\( Z = 400x + 200y \), which is to be minimized under constraints
\( 200x + 80y \geq 1200 \)
\( \implies \) \( 5x + 2y \geq 30 \)
\( 400x + 200y \leq 3000 \)
\( \implies \) \( 2x + y \leq 15 \)
\( x \leq y, x \geq 0, y \geq 0 \)

Linear Programming (LPP)

Question 1. Two tailors A and B earn Rs.150 and Rs.200 per day respectively. 'A' can stitch 6 shirts and 4 pants per day. While 'B' can stitch 10 shirts and 4 pants per day. Form LPP to minimize the labour cost to produce at least 60 shirts and 32 pants, and solve it.
Answer:
Let \( x \) and \( y \) denote the number of days tailor A and tailor B work, respectively. Let \( Z \) represent the total expenditure on wages.
The linear programming problem can be mathematically formulated as follows:
\[ \text{Minimize } Z = 150x + 200y \]
Subject to the constraints:
1) Shirt production constraint (at least 60 shirts):
\[ 6x + 10y \ge 60 \]
2) Pant production constraint (at least 32 pants):
\[ 4x + 4y \ge 32 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
To find the feasible region, we determine the boundary points for each constraint line:
- For the first constraint \( 6x + 10y = 60 \), the intersecting coordinates are \( (10, 0) \) and \( (0, 6) \). Since the inequality is \( \ge \), the shaded region lies away from the origin.
- For the second constraint \( 4x + 4y = 32 \), the intersecting coordinates are \( (8, 0) \) and \( (0, 8) \). The shaded region lies away from the origin.
By plotting these lines, we obtain an open, unbounded feasible region with corner points at \( A(10, 0) \), \( B(5, 3) \), and \( C(0, 8) \). The intersection point \( B(5, 3) \) is found by solving the two boundary equations simultaneously.

We now compute the objective cost function at each of the corner points:
- At \( A(10, 0) \): \( Z = 150(10) + 200(0) = 1500 \)
- At \( B(5, 3) \): \( Z = 150(5) + 200(3) = 750 + 600 = 1350 \)
- At \( C(0, 8) \): \( Z = 150(0) + 200(8) = 1600 \)

The minimum wage cost of Rs. 1350 is achieved when tailor A is hired for 5 days and tailor B is hired for 3 days.
In simple words: To complete the required stitching of shirts and pants at the lowest possible cost, tailor A should work for 5 days and tailor B should work for 3 days. This will cost a total of Rs. 1350.

Exam Tip: For minimization problems with an unbounded region, remember that the minimum value obtained at the corner point is valid because the coefficients in the objective function are positive.

Question 2. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs.5760 to invest and has space for at most 20 items. A fan costs him Rs. 360 and a sewing machine Rs.240. He can sell a fan at a profit of Rs.22 and a sewing machine at a profit of Rs. 18. How should he invest his money in order to maximize his profit.
Answer:
Let \( x \) and \( y \) be the quantity of fans and sewing machines purchased by the dealer, respectively. Let \( Z \) represent the total profit.
The problem can be formulated as follows:
\[ \text{Maximize } Z = 22x + 18y \]
Subject to the constraints:
1) Financial investment limit (at most Rs. 5760):
\[ 360x + 240y \le 5760 \implies 3x + 2y \le 48 \]
2) Storage capacity restriction (at most 20 items):
\[ x + y \le 20 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
By determining the intercepts, we can plot the boundary lines:
- For \( 3x + 2y = 48 \), the boundary line connects \( (16, 0) \) and \( (0, 24) \).
- For \( x + y = 20 \), the boundary line connects \( (20, 0) \) and \( (0, 20) \).
The intersection of these constraints forms a closed, bounded feasible region with corner points at \( O(0,0) \), \( A(16, 0) \), \( B(8, 12) \), and \( C(0, 20) \).

We now compute the total profit \( Z \) at each corner vertex of this feasible region:
- At \( A(16, 0) \): \( Z = 22(16) + 18(0) = 352 \)
- At \( B(8, 12) \): \( Z = 22(8) + 18(12) = 176 + 216 = 392 \)
- At \( C(0, 20) \): \( Z = 22(0) + 18(20) = 360 \)

The maximum profit is Rs. 392, which occurs when the dealer purchases 8 fans and 12 sewing machines.
In simple words: To get the highest profit, the shopkeeper should buy 8 fans and 12 sewing machines. This setup fits within the Rs. 5760 budget and uses the available space perfectly, bringing in Rs. 392 in profit.

Exam Tip: Simplify the numbers in equations (like dividing the entire cost inequality by 120) before sketching lines. It saves time and prevents calculation mistakes during the graph-plotting stage.

Question 3. A manufacturer of a line of patent medicines is preparing a production plan on medicines A and B. There are sufficient ingredients available to make 20,000 bottles of A and 40,000 bottles of B but there are only 45,000 bottles in to which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to fill 1000 bottles of A, it takes one hour to prepare enough material to fill 1000 bottle of B and there are 66 hours available for this operation. The profit is Rs. 8 per bottle for A and Rs. 7 per bottle for B formulate this problem as LPP and solve it.
Answer:
Let \( x \) denote the number of bottles of medicine A produced, and let \( y \) denote the number of bottles of medicine B produced. Let \( Z \) represent the total profit.
We formulate the linear programming problem as follows:
\[ \text{Maximize } Z = 8x + 7y \]
Subject to the constraints:
1) Combined bottle capacity constraint:
\[ x + y \le 45000 \]
2) Ingredient limitation for medicine A:
\[ x \le 20000 \]
3) Ingredient limitation for medicine B:
\[ y \le 40000 \]
4) Preparation time constraint (with 66 hours available, taking 3 hours per 1000 bottles of A and 1 hour per 1000 bottles of B):
\[ \frac{3x}{1000} + \frac{1y}{1000} \le 66 \implies 3x + y \le 66000 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these boundaries, we identify the corner points of our feasible region: \( A(20000, 0) \), \( B(20000, 6000) \), \( C(10500, 34500) \), \( D(5000, 40000) \), and \( E(0, 40000) \).

We now compute the total profit \( Z \) at each corner vertex:
- At \( A(20000, 0) \): \( Z = 8(20000) + 7(0) = 1,60,000 \)
- At \( B(20000, 6000) \): \( Z = 8(20000) + 7(6000) = 1,60,000 + 42,000 = 2,02,000 \)
- At \( C(10500, 34500) \): \( Z = 8(10500) + 7(34500) = 84,000 + 2,41,500 = 3,25,500 \)
- At \( D(5000, 40000) \): \( Z = 8(5000) + 7(40000) = 40,000 + 2,80,000 = 3,20,000 \)
- At \( E(0, 40000) \): \( Z = 8(0) + 7(40000) = 2,80,000 \)

The highest profit of Rs. 3,25,500 occurs at the intersection point \( C \).
Therefore, the manufacturer should produce 10,500 bottles of medicine A and 34,500 bottles of medicine B to get the maximum profit of Rs. 3,25,500.
In simple words: To make the absolute maximum profit, the medicine company needs to fill 10,500 bottles of A and 34,500 bottles of B. This keeps the total profit at a maximum of Rs. 3,25,500.

Exam Tip: Pay special attention to constraints that are defined in terms of thousands of items (like hours per 1000 bottles). Converting these fractions to a standard linear form such as \( 3x + y \le 66000 \) prevents algebraic mistakes.

Question 4. Every gram of wheat provides 0.1 𝑔𝑚 of proteins and 0.25 𝑔𝑚 of carbohydrates. The corresponding values of rice are 0.05 𝑔𝑚 and 0.5 𝑔𝑚 respectively. Wheat cost 𝑅𝑠. 4 𝑝𝑒𝑟 𝑘𝑔 and rice Rs. 6 per kg. The minimum daily requirement of proteins and carbohydrates are 50𝑔𝑚 and 200𝑔𝑚 respectively. In what quantities should wheat and rice be mixed to provide minimum daily requirements of proteins and carbohydrates at minimum cost.
Answer:
Let \( x \) represent the mass (in grams) of wheat and \( y \) represent the mass (in grams) of rice mixed in the daily diet.
Since the costs are given per kilogram, we convert the variables to kilograms or adjust the coefficients. The cost per gram of wheat is \( \frac{4}{1000} \) and for rice is \( \frac{6}{1000} \). Our objective function is:
\[ \text{Minimize } Z = \frac{4x}{1000} + \frac{6y}{1000} \]
Subject to the nutritional constraints:
1) Daily protein constraint (at least 50 gm):
\[ 0.1x + 0.05y \ge 50 \implies 10x + 5y \ge 5000 \implies 2x + y \ge 1000 \]
2) Daily carbohydrate constraint (at least 200 gm):
\[ 0.25x + 0.5y \ge 200 \implies 25x + 50y \ge 20000 \implies x + 2y \ge 800 \]
The non-negativity constraints are \( x \ge 0, y \ge 0 \).
The feasible region is open and unbounded away from the origin, with corner points at \( A(800, 0) \), \( B(400, 200) \), and \( C(0, 1000) \).

We now evaluate the cost function \( Z \) at each corner vertex:
- At \( A(800, 0) \): \( Z = \frac{4(800)}{1000} + 0 = 3.2 \)
- At \( B(400, 200) \): \( Z = \frac{4(400)}{1000} + \frac{6(200)}{1000} = 1.6 + 1.2 = 2.8 \)
- At \( C(0, 1000) \): \( Z = 0 + \frac{6(1000)}{1000} = 6 \)

The minimum daily diet cost is Rs. 2.8, which is achieved with a mixture of 400 gm of wheat and 200 gm of rice.
In simple words: To get enough daily protein and carbs at the lowest price, you should eat 400 grams of wheat and 200 grams of rice every day. This daily diet will cost Rs. 2.8.

Exam Tip: Since variables are in grams but costs are given per kilogram, you must convert the units to match before formulating your objective function to avoid decimal placing errors.

Question 5. An oil company has two depots A and B with capacities of 7000 liters and 4000 liters respectively. The company is to supply oil to three petrol pumps D, E, and F whose requirements are 4500, 3000 and 3500 liters respectively. The distance in (km) between the depots and patrol pumps is given in table :-

Assuming that the transportation cost per km is Rs. 1.00 per ten liters. How should the delivery be scheduled to minimize the transportation cost.
Answer:
Let \( x \) liters of oil be transported from depot A to petrol pump D, and let \( y \) liters be transported to petrol pump E.
Since the total capacity of depot A is 7000 liters, the quantity transported from A to F must be \( 7000 - x - y \) liters.
Now, the remaining demands of each petrol pump are supplied by depot B:
- Transported from B to D: \( 4500 - x \) liters
- Transported from B to E: \( 3000 - y \) liters
- Transported from B to F: \( 3500 - (7000 - x - y) = x + y - 3500 \) liters
Because these physical quantities must be non-negative, we get the following system of linear inequalities:
\[ x \ge 0, \quad y \ge 0 \]
\[ 7000 - x - y \ge 0 \implies x + y \le 7000 \]
\[ 4500 - x \ge 0 \implies x \le 4500 \]
\[ 3000 - y \ge 0 \implies y \le 3000 \]
\[ x + y - 3500 \ge 0 \implies x + y \ge 3500 \]
The cost rate of transportation is Rs. 1.00 per 10 liters per km (which is equivalent to Rs. \( \frac{1}{10} \) per liter per km). The total shipping cost function \( Z \) is calculated as:
\[ Z = \frac{1}{10} [ 7x + 6y + 3(7000 - x - y) + 3(4500 - x) + 4(3000 - y) + 2(x + y - 3500) ] \]
\[ Z = \frac{1}{10} (3x + y + 39500) \]
We now graph these constraints to locate the bounded feasible region and evaluate \( Z \) at its corner points:
- At \( A(3000, 500) \): \( Z = \frac{1}{10} (3(3000) + 500 + 39500) = 4850 \)
- At \( B(4500, 0) \): \( Z = \frac{1}{10} (3(4500) + 0 + 39500) = 5300 \)
- At \( C(4500, 2500) \): \( Z = \frac{1}{10} (3(4500) + 2500 + 39500) = 5550 \)
- At \( D(4000, 3000) \): \( Z = \frac{1}{10} (3(4000) + 3000 + 39500) = 5450 \)
- At \( E(500, 3000) \): \( Z = \frac{1}{10} (3(500) + 3000 + 39500) = 4400 \)

The minimum transportation cost is Rs. 4400, which is attained at corner point \( E(500, 3000) \).
Thus, the optimal transportation schedule is as follows:
- From Depot A: 500 liters to D, 3000 liters to E, and 3500 liters to F.
- From Depot B: 4000 liters to D, 0 liters to E, and 0 liters to F.
In simple words: To pay the least amount in shipping fees, the oil company should send 500 liters to D, 3000 liters to E, and 3500 liters to F from Depot A. The remaining 4000 liters for D should be delivered from Depot B. This keeps the total cost at Rs. 4400.

Exam Tip: For complex transportation problems with multiple locations, create a small network flow diagram to define the variables \( x \) and \( y \). This lets you reduce the problem to two variables and solve it quickly using a 2D graph.

Question 6. If a young man drives his vehicle at 25 km/h, he has to spend 𝑅𝑠. 2 𝑝𝑒𝑟 𝑘𝑚 on petrol. If he drives it at a faster speed of 40 km/h, the petrol cost increases to Rs. 5 per km. He has 𝑅𝑠. 100 to spend on petrol and travel with one hour. Express this on LPP and solve the same.
Answer:
Let \( x \) denote the distance (in km) traveled at the speed of 25 km/h, and let \( y \) denote the distance (in km) traveled at the speed of 40 km/h. Let \( Z \) be the total distance traveled.
The LPP can be formulated as follows:
\[ \text{Maximize } Z = x + y \]
Subject to the constraints:
1) Budget constraint on petrol cost (maximum Rs. 100):
\[ 2x + 5y \le 100 \]
2) Time constraint (maximum 1 hour, using \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)):
\[ \frac{x}{25} + \frac{y}{40} \le 1 \implies 8x + 5y \le 200 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we locate the feasible region bounded by the corner points \( O(0,0) \), \( A(25, 0) \), \( B\left(\frac{50}{3}, \frac{40}{3}\right) \), and \( C(0, 20) \):

We now compute the objective distance \( Z \) at each vertex:
- At \( A(25, 0) \): \( Z = 25 + 0 = 25 \)
- At \( B\left(\frac{50}{3}, \frac{40}{3}\right) \): \( Z = \frac{50}{3} + \frac{40}{3} = \frac{90}{3} = 30 \) (Maximum)
- At \( C(0, 20) \): \( Z = 0 + 20 = 20 \)

The maximum distance traveled is 30 km, which is achieved when driving \( \frac{50}{3} \) km at 25 km/h and \( \frac{40}{3} \) km at 40 km/h.
In simple words: To travel the longest possible distance within one hour and stay inside the Rs. 100 budget, the young man must ride about 16.7 km at 25 km/h and 13.3 km at 40 km/h. This gives a total distance of 30 km.

Exam Tip: The time constraint requires combining distance and speed to formulate the expression. Always keep track of units to make sure your constraint equation has time as its final dimension on both sides.

Question 7. A company makes two kinds of leather belts A and B. But A is high quality belt and B is of lower quality. The respective profit are Rs. 4 and Rs 3 per belt. Each belt of type A requires twice as much time as a belt of type B and if all belts of types B the company could make 1000 belts per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 buckles per day are available. There are only 700 buckles available for belt B. What should be the daily production of each type of belt to maximum the profit.
Answer:
Let \( x \) be the number of type A belts and \( y \) be the number of type B belts produced daily. Let \( Z \) denote the daily profit.
The linear programming model is formulated as follows:
\[ \text{Maximize } Z = 4x + 3y \]
Subject to the constraints:
1) Production time limitation (since B takes \( t \) hours and A takes \( 2t \) hours, and 1000 belts of B can be made per day):
\[ 2x + y \le 1000 \]
2) Raw leather availability constraint (at most 800 belts total):
\[ x + y \le 800 \]
3) Buckle availability for belt A (at most 400):
\[ x \le 400 \]
4) Buckle availability for belt B (at most 700):
\[ y \le 700 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities on a graph, we locate the corner points of the closed feasible region: \( A(400, 0) \), \( B(400, 200) \), \( C(200, 600) \), \( D(100, 700) \), and \( E(0, 700) \).

We now compute the profit \( Z \) at each corner vertex:
- At \( A(400, 0) \): \( Z = 4(400) + 3(0) = 1600 \)
- At \( B(400, 200) \): \( Z = 4(400) + 3(200) = 2200 \)
- At \( C(200, 600) \): \( Z = 4(200) + 3(600) = 2600 \) (Maximum)
- At \( D(100, 700) \): \( Z = 4(100) + 3(700) = 2500 \)
- At \( E(0, 700) \): \( Z = 4(0) + 3(700) = 2100 \)

The optimal profit of Rs. 2600 is achieved when producing 200 high-quality belts (kind A) and 600 lower-quality belts (kind B) per day.
In simple words: To earn the absolute highest daily profit of Rs. 2600, the factory must manufacture 200 belts of Type A and 600 belts of Type B. This plan fits all time, leather, and buckle restrictions.

Exam Tip: When a problem has multiple resource constraints (like raw leather, time, and two different buckle limits), make sure to draw every single boundary line. Dropping one constraint can incorrectly enlarge your feasible region.

Question 8. A toy manufacturer produces two types of dolls ; a basic version doll A and deluxe version doll B. each doll of type B takes twice as long to produce as one doll of type A. The company have time to make a maximum of 2000 dolls of type A per day, the supply of plastic is sufficient to produce 1500 dolls per day. The deluxe version B requires a fancy dress of which there are only 600 per day available. If the company makes profit of Rs 3 and Rs 5 per doll on A & B respectively. How many of each should be produced to maximize the profit ?
Answer:
Let \( x \) and \( y \) represent the daily production quantities of basic type A dolls and deluxe type B dolls, respectively. Let \( Z \) denote the daily profit.
The LPP is formulated to maximize returns:
\[ \text{Maximize } Z = 3x + 5y \]
Subject to the constraints:
1) Production time constraint (since B takes twice as long as A, and max capacity is 2000 dolls of A):
\[ x + 2y \le 2000 \]
2) Raw plastic availability constraint (at most 1500 dolls total):
\[ x + y \le 1500 \]
3) Fancy dress availability for doll B (at most 600):
\[ y \le 600 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we locate the bounded feasible region with corner points at \( O(0,0) \), \( (1500,0) \), \( P(1000,500) \), \( (800,600) \), and \( (0,600) \).

We now compute the objective profit \( Z \) at each corner vertex:
- At \( (1500, 0) \): \( Z = 3(1500) + 5(0) = 4500 \)
- At \( P(1000, 500) \): \( Z = 3(1000) + 5(500) = 3000 + 2500 = 5500 \) (Maximum)
- At \( (800, 600) \): \( Z = 3(800) + 5(600) = 2400 + 3000 = 5400 \)
- At \( (0, 600) \): \( Z = 3(0) + 5(600) = 3000 \)

The maximum profit is Rs. 5500, which occurs when producing 1000 basic dolls (type A) and 500 deluxe dolls (type B) daily.
In simple words: To get the highest daily profit of Rs. 5500, the toy company should produce exactly 1000 basic dolls and 500 deluxe dolls. This keeps the factory running optimally within the material and time limits.

Exam Tip: Be careful when formulating rates like "takes twice as long". Writing the time constraint in terms of standard doll units (\( x + 2y \le 2000 \)) is the easiest way to prevent algebraic mistakes.

Question 9. Two godown A and B have grain storage capacity of 100 quintals and 50 quintals respectively. They supply to 3 ration shops D, E and F , whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintal from the god0wn to the shop are given in table

how should the supplies be transported in order that the transportation cost is minimum ?
Answer:
Let \( x \) and \( y \) represent the amount of grain (in quintals) transported from godown A to ration shops D and E, respectively. Since godown A has a total capacity of 100 quintals, the amount transported from A to F must be \( 100 - (x+y) \) quintals.
The remaining requirements of the shops are supplied by godown B:
- To D: \( 60 - x \) quintals
- To E: \( 50 - y \) quintals
- To F: \( 40 - [100 - (x+y)] = x + y - 60 \) quintals
Because all these physical transport quantities must be non-negative, we get:
\[ x \ge 0, \quad y \ge 0 \]
\[ 100 - (x+y) \ge 0 \implies x + y \le 100 \]
\[ 60 - x \ge 0 \implies x \le 60 \]
\[ 50 - y \ge 0 \implies y \le 50 \]
\[ x + y - 60 \ge 0 \implies x + y \ge 60 \]
The total transportation cost \( Z \) is formulated as:
\[ Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60) \]
\[ Z = 2.5x + 1.5y + 410 \]
Our goal is to minimize this objective function:
\[ \text{Minimize } Z = 2.5x + 1.5y + 410 \]
Plotting these inequalities on a graph, we locate the corner points of our closed feasible region: \( A(10, 50) \), \( B(50, 50) \), \( C(60, 40) \), and \( D(60, 0) \).

We now compute the total cost \( Z \) at each corner vertex:
- At \( A(10, 50) \): \( Z = 2.5(10) + 1.5(50) + 410 = 25 + 75 + 410 = 510 \) (Minimum)
- At \( B(50, 50) \): \( Z = 2.5(50) + 1.5(50) + 410 = 125 + 75 + 410 = 610 \)
- At \( C(60, 40) \): \( Z = 2.5(60) + 1.5(40) + 410 = 150 + 60 + 410 = 620 \)
- At \( D(60, 0) \): \( Z = 2.5(60) + 1.5(0) + 410 = 150 + 0 + 410 = 560 \)

The minimum cost is Rs. 510, which occurs at \( A(10, 50) \).
Therefore, the optimal transportation schedule is as follows:
- From Godown A: 10 quintals to D, 50 quintals to E, and 40 quintals to F.
- From Godown B: 50 quintals to D, 0 quintals to E, and 0 quintals to F.
In simple words: To pay the absolute lowest transportation fee of Rs. 510, the company should send 10 quintals of grain to D, 50 to E, and 40 to F from Godown A. The remaining 50 quintals for D should be delivered from Godown B.

Exam Tip: For complex transportation problems, always establish all the variables in terms of \( x \) and \( y \) using a simple network diagram. This reduces the problem to a standard two-variable LPP that can easily be solved graphically.

Question 10. One kind of cake requires 300 𝑔𝑚 of flour and 15 𝑔𝑚 of fat, another kind of cake requires 150 𝑔𝑚 of flour and 30 𝑔𝑚 of fat. Find the maximum number of cakes which can be made from 7.5 𝑘𝑔 of flour and 600 𝑔𝑚 of fat, assuming there is no shortage. Make it as on LPP and solve it.
Answer:
Let \( x \) and \( y \) denote the number of cakes of type I and type II made, respectively. We construct the following table from the given conditions:

We want to maximize the total number of cakes produced:
\[ \text{Maximize } Z = x + y \]
Subject to the following constraints:
1) Flour capacity restriction (maximum 7500 gm):
\[ 300x + 150y \le 7500 \implies 2x + y \le 50 \]
2) Fat capacity restriction (maximum 600 gm):
\[ 15x + 30y \le 600 \implies x + 2y \le 40 \]
Additionally, we have non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
The boundary lines intersect to form a closed feasible region with corner points at \( O(0, 0) \), \( A(25, 0) \), \( B(0, 20) \), and \( C(20, 10) \).

We now compute the objective function \( Z = x + y \) at each corner vertex:
- At \( O(0, 0) \): \( Z = 0 + 0 = 0 \)
- At \( A(25, 0) \): \( Z = 25 + 0 = 25 \)
- At \( B(0, 20) \): \( Z = 0 + 20 = 20 \)
- At \( C(20, 10) \): \( Z = 20 + 10 = 30 \) (Maximum)

The maximum value of \( Z \) is 30, achieved at \( C(20, 10) \).
Therefore, the maximum number of cakes that can be made is 30, consisting of 20 cakes of type I and 10 cakes of type II.
In simple words: To make the most cakes possible using the available flour and fat, the baker should bake exactly 20 cakes of type I and 10 cakes of type II. This makes a total of 30 cakes.

Exam Tip: Always make sure to write down units carefully before starting. Converting 7.5 kg of flour directly into 7500 gm of flour ensures all your constraint equations have matching, consistent units.