CBSE Class 12 Mathematics Linear Programming Assignment Set 02

Points to Remember

• Linear programming is a mathematical technique used to find the optimal (highest or lowest) value of a linear objective function, subject to a set of given linear constraints.
• Objective Functions: The linear function \( z = ax + by \) (where \( a \) and \( b \) are constants) that needs to be maximized or minimized is known as the linear objective function.
• Constraints: These are the linear inequalities, equations, or limitations imposed on the variables in a linear programming problem.
• Feasible Region: The common area or set of points that satisfies all the given constraints simultaneously.
• To Find Feasible Region: Plot the boundary lines of all linear inequations on a coordinate plane and shade the overlapping region that satisfies all constraints.
• Feasible Solutions: Any point located inside or on the boundary of the feasible region represents a valid feasible solution.
• Optimal Feasible Solution: The specific feasible solution that yields the maximum or minimum value of the objective function.

Question 1. Solve the following L.P.P. graphically
Minimise and maximise \( z = 3x + 9y \)
Subject to the constraints \( x + 3y \le 60 \)
\( x + y \ge 10 \)
\( x \le y \)
\( x \ge 0, y \ge 0 \)
Answer: To solve the given linear programming problem, we first plot the boundary lines for each constraint:
1. The line \( x + 3y = 60 \) passes through the points \( (0, 20) \) and \( (60, 0) \). Since the inequality is \( x + 3y \le 60 \), the region lies towards the origin.
2. The line \( x + y = 10 \) passes through the points \( (0, 10) \) and \( (10, 0) \). Since the inequality is \( x + y \ge 10 \), the region lies away from the origin.
3. The line \( y = x \) passes through \( (0, 0) \) and \( (10, 10) \). The inequality \( x \le y \) represents the region above this line.
The common shaded region representing the feasible region is bounded by the following corner points:
- \( A(0, 10) \)
- \( B(5, 5) \) (intersection of \( x + y = 10 \) and \( y = x \))
- \( C(15, 15) \) (intersection of \( x + 3y = 60 \) and \( y = x \))
- \( D(0, 20) \)
Now, let us evaluate the objective function \( z = 3x + 9y \) at each corner point:
- At \( A(0, 10) \): \( z = 3(0) + 9(10) = 90 \)
- At \( B(5, 5) \): \( z = 3(5) + 9(5) = 15 + 45 = 60 \)
- At \( C(15, 15) \): \( z = 3(15) + 9(15) = 45 + 135 = 180 \)
- At \( D(0, 20) \): \( z = 3(0) + 9(20) = 180 \)
Comparing these values:
\( \implies \) The minimum value of \( z \) is \( 60 \) at point \( (5, 5) \).
\( \implies \) The maximum value of \( z \) is \( 180 \) at both \( (15, 15) \) and \( (0, 20) \), which means the maximum is achieved at all points on the line segment joining these two points.
In simple words: First find the corners of the shaded region where all your boundary lines intersect. Calculate the value of the equation at each corner point to find the lowest and highest results.

Exam Tip: When the maximum value is identical at two corner points, specify that the maximum occurs at all points along the line segment connecting those two points, as this is a key conceptual point that carries marks.

Question 2. Determine graphically the minimum value of the objective function \( z = - 50x + 20 y \), subject to the constraints
\( 2x - y \ge - 5 \)
\( 3x + y \ge 3 \)
\( 2x - 3y \le 12 \)
\( x \ge 0, y \ge 0 \)
Answer: Let us plot the lines representing each of the given constraint equations:
1. The line \( 2x - y = -5 \) passes through the points \( (0, 5) \) and \( (-2.5, 0) \).
2. The line \( 3x + y = 3 \) passes through \( (0, 3) \) and \( (1, 0) \).
3. The line \( 2x - 3y = 12 \) passes through \( (6, 0) \) and \( (0, -4) \).
By shading the regions corresponding to each inequality in the first quadrant (\( x \ge 0, y \ge 0 \)), we obtain an unbounded feasible region with the following corner points:
- \( A(0, 5) \)
- \( B(0, 3) \)
- \( C(1, 0) \)
- \( D(6, 0) \)
Let us evaluate the objective function \( z = -50x + 20y \) at these corner points:
- At \( A(0, 5) \): \( z = -50(0) + 20(5) = 100 \)
- At \( B(0, 3) \): \( z = -50(0) + 20(3) = 60 \)
- At \( C(1, 0) \): \( z = -50(1) + 20(0) = -50 \)
- At \( D(6, 0) \): \( z = -50(6) + 20(0) = -300 \)
Since the feasible region is unbounded, we must check if there is any other value in the region such that \( z < -300 \). We graph the inequality \( -50x + 20y < -300 \implies -5x + 2y < -30 \).
Since this open half-plane has no common points with our feasible region, the minimum value is confirmed.
\( \implies \) The minimum value of \( z \) is \( -300 \) at the point \( (6, 0) \).
In simple words: Plot all the boundary equations and identify the unbounded region. Find the values at the corner points, and verify that no other points in the unbounded region can yield a smaller value.

Exam Tip: For unbounded regions, do not declare the minimum value from the corner points directly. You must always plot the inequality \( ax + by < M \) and state that there are no common points to get full marks.

Question 3. 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. Formulate the above L.P.P. mathematically and hence solve it to minimise the labour cost to produce at least 60 shirts and 32 pants.
Answer: Let Tailor A work for \( x \) days and Tailor B work for \( y \) days. The mathematical formulation of the given problem is:
Minimize the total labour cost:
\( Z = 150x + 200y \)
Subject to the constraints:
- For shirts: \( 6x + 10y \ge 60 \implies 3x + 5y \ge 30 \)
- For pants: \( 4x + 4y \ge 32 \implies x + y \ge 8 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us find the corner points of the unbounded feasible region:
- \( A(0, 8) \)
- \( B(5, 3) \) (the intersection of the lines \( 3x + 5y = 30 \) and \( x + y = 8 \))
- \( C(10, 0) \)
Let us evaluate the objective function \( Z = 150x + 200y \) at each corner point:
- At \( A(0, 8) \): \( Z = 150(0) + 200(8) = 1600 \)
- At \( B(5, 3) \): \( Z = 150(5) + 200(3) = 750 + 600 = 1350 \)
- At \( C(10, 0) \): \( Z = 150(10) + 200(0) = 1500 \)
Since the feasible region is unbounded, we verify the minimum value by plotting the inequality \( 150x + 200y < 1350 \implies 3x + 4y < 27 \). This open half-plane has no common points with the feasible region.
\( \implies \) The minimum labour cost is Rs. 1350, which is achieved when Tailor A works for 5 days and Tailor B works for 3 days.
In simple words: Represent the number of days worked by the tailors as \( x \) and \( y \). Formulate inequalities based on the required shirts and pants, then find the corner points of the feasible region to minimize the cost.

Exam Tip: Be sure to formulate the inequalities with a "greater-than-or-equal-to" (\( \ge \)) sign, because the problem specifies "at least" 60 shirts and 32 pants.

Question 4. There are two types of fertilisers A and B. A consists of 10% nitrogen and 6% phosphoric acid and B consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for his crop. If A costs Rs. 61 kg and B costs Rs. 51 kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at minimum cost. What is the minimum cost?
Answer: Let \( x \) kg of fertiliser A and \( y \) kg of fertiliser B be used.
The constraints based on the nutrient requirements are:
- For Nitrogen: \( 10\% \text{ of } x + 5\% \text{ of } y \ge 14 \implies 0.1x + 0.05y \ge 14 \implies 2x + y \ge 280 \)
- For Phosphoric Acid: \( 6\% \text{ of } x + 10\% \text{ of } y \ge 14 \implies 0.06x + 0.1y \ge 14 \implies 3x + 5y \ge 700 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
The corner points of the unbounded feasible region are:
- \( P(0, 280) \)
- \( Q(100, 80) \) (intersection of \( 2x + y = 280 \) and \( 3x + 5y = 700 \))
- \( R(233.33, 0) \)
Note: The printed costs in the question (Rs. 61 kg and Rs. 51 kg) appear to have a typographical error (representing Rs. 6 per kg and Rs. 5 per kg). We evaluate both cases below:
Case I: Using standard corrected costs of Rs. 6 per kg and Rs. 5 per kg
Objective function: \( Z = 6x + 5y \)
- At \( P(0, 280) \): \( Z = 6(0) + 5(280) = 1400 \)
- At \( Q(100, 80) \): \( Z = 6(100) + 5(80) = 1000 \) (Minimum)
- At \( R(700/3, 0) \): \( Z = 6(700/3) = 1400 \)
\( \implies \) The minimum cost is Rs. 1000 when 100 kg of fertiliser A and 80 kg of fertiliser B are used.
Case II: Using verbatim costs from the text (Rs. 61 per kg and Rs. 51 per kg)
Objective function: \( Z = 61x + 51y \)
- At \( P(0, 280) \): \( Z = 61(0) + 51(280) = 14280 \)
- At \( Q(100, 80) \): \( Z = 61(100) + 51(80) = 6100 + 4080 = 10180 \) (Minimum)
- At \( R(700/3, 0) \): \( Z \approx 14233.33 \)
\( \implies \) The minimum cost is Rs. 10180 when 100 kg of fertiliser A and 80 kg of fertiliser B are used.
In simple words: Set up equations using the percentages of nutrients required. Find the corner points of the shaded region and use the cost prices to calculate the most economical combination.

Exam Tip: Tabulate the given data (fertiliser types, nitrogen content, phosphoric acid content, and costs) as a first step to make the formulation of constraint equations easier and mistake-free.

Question 5. A man has Rs. 1500 to purchase two types of shares of two different companies \( S_1 \) and \( S_2 \). Market price of one share of \( S_1 \) is Rs 180 and \( S_2 \) is Rs. 120. He wishes to purchase a maximum of ten shares only. If one share of type \( S_1 \) gives a yield of Rs. 11 and of type \( S_2 \) yields Rs. 8 then how much shares of each type must be purchased to get maximum profit? And what will be the maximum profit?
Answer: Let \( x \) be the number of shares of company \( S_1 \) and \( y \) be the number of shares of company \( S_2 \).
The objective function to maximize total profit is:
\( Z = 11x + 8y \)
Subject to the constraints:
- Budget: \( 180x + 120y \le 1500 \implies 3x + 2y \le 25 \)
- Share limit: \( x + y \le 10 \)
- Non-negativity and integer constraint: \( x \ge 0, y \ge 0, \; x, y \in \mathbb{Z} \)
The corner points of the bounded feasible region are:
- \( A(0, 10) \)
- \( B(5, 5) \) (intersection of \( 3x + 2y = 25 \) and \( x + y = 10 \))
- \( C(8.33, 0) \implies \) For integer constraints, we consider \( (8, 0) \)
Let us evaluate the profit \( Z \) at these points:
- At \( A(0, 10) \): \( Z = 11(0) + 8(10) = 80 \)
- At \( B(5, 5) \): \( Z = 11(5) + 8(5) = 55 + 40 = 95 \)
- At \( C(8, 0) \): \( Z = 11(8) + 8(0) = 88 \)
\( \implies \) To maximize his profit, the man should purchase 5 shares of company \( S_1 \) and 5 shares of company \( S_2 \). The maximum profit is Rs. 95.
In simple words: Use \( x \) and \( y \) for the quantities of both shares. Set up inequalities for the budget and total count limits, check the profits at all corners, and pick the highest result.

Exam Tip: Since shares must be purchased in whole numbers, make sure your optimal corner points contain only integer values.

Question 6. A company manufactures two types of lamps say A and B. Both lamps go through a cutter and then a finisher. Lamp A requires 2 hours of the cutter’s time and 1 hours of the finisher’s time. Lamp B requires 1 hour of cutter’s and 2 hours of finisher’s time. The cutter has 100 hours and finishers has 80 hours of time available each month. Profit on one lamp A is Rs. 7.00 and on one lamp B is Rs. 13.00. Assuming that he can sell all that he produces, how many of each type of lamps should be manufactured to obtain maximum profit?
Answer: Let \( x \) be the number of lamps of type A produced, and \( y \) be the number of lamps of type B produced.
The objective function to maximize profit is:
\( Z = 7x + 13y \)
Subject to the constraints:
- Cutter time: \( 2x + y \le 100 \)
- Finisher time: \( x + 2y \le 80 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us determine the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( A(50, 0) \) (intersection of \( 2x + y = 100 \) and \( y = 0 \))
- \( B(40, 30) \) (intersection of \( 2x + y = 100 \) and \( x + 2y = 80 \))
- \( C(0, 40) \) (intersection of \( x + 2y = 80 \) and \( x = 0 \))
Let us evaluate \( Z \) at each corner point:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(50, 0) \): \( Z = 7(50) + 13(0) = 350 \)
- At \( B(40, 30) \): \( Z = 7(40) + 13(30) = 280 + 390 = 670 \)
- At \( C(0, 40) \): \( Z = 7(0) + 13(40) = 520 \)
Comparing these values, the maximum profit is Rs. 670.
\( \implies \) The company should produce 40 lamps of type A and 30 lamps of type B to maximize profit.
In simple words: Represent the two lamps as \( x \) and \( y \). Establish inequalities using cutter and finisher hour limits, then check the profit at the intersections to find the best mix.

Exam Tip: Clearly draw the axes and identify the shaded feasible region, labeling all four corner vertices with their calculated coordinates.

Question 7. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs. 5760 to invest and has space for almost 20 items. A fan and sewing machine cost Rs. 360 and Rs. 240 respectively. He can sell a fan at a profit of Rs. 22 and sewing machine at a profit of Rs. 18. Assuming that he can sell whatever he buys, how should he invest his money to maximise his profit?
Answer: Let \( x \) be the number of fans purchased, and \( y \) be the number of sewing machines purchased.
The objective function to maximize profit is:
\( Z = 22x + 18y \)
Subject to the constraints:
- Investment constraint: \( 360x + 240y \le 5760 \implies 3x + 2y \le 48 \)
- Space constraint: \( x + y \le 20 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us determine the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( A(16, 0) \) (from \( 3x = 48 \))
- \( B(8, 12) \) (intersection of \( 3x + 2y = 48 \) and \( x + y = 20 \))
- \( C(0, 20) \) (from \( x + y = 20 \))
Let us evaluate the profit \( Z \) at each corner point:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(16, 0) \): \( Z = 22(16) + 18(0) = 352 \)
- At \( B(8, 12) \): \( Z = 22(8) + 18(12) = 176 + 216 = 392 \)
- At \( C(0, 20) \): \( Z = 22(0) + 18(20) = 360 \)
Comparing these values, the maximum profit is Rs. 392.
\( \implies \) The dealer should invest in 8 fans and 12 sewing machines to get the maximum profit of Rs. 392.
In simple words: Represent fans and sewing machines as variables. Set up constraints for cost and storage space, plot them to find the vertices of the common region, and solve for the highest profit.

Exam Tip: Be sure to simplify the budget inequality by dividing by the common divisor (120) before plotting to keep your coordinate values small and simple.

Question 8. If a young man rides his motorcycle at 25 km/h, he has to spend Rs. 2 per km on petrol. If he rides 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 wishes to find the maximum distance he can travel within one hour. Express this as L.P.P. and then solve it graphically.
Answer: Let the young man ride \( x \) km at a speed of 25 km/h and \( y \) km at a speed of 40 km/h.
The objective function to maximize total distance is:
\( Z = x + y \)
Subject to the constraints:
- Petrol expense: \( 2x + 5y \le 100 \)
- Time limit: \( \frac{x}{25} + \frac{y}{40} \le 1 \implies 8x + 5y \le 200 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us find the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( A(25, 0) \)
- \( B\left(\frac{50}{3}, \frac{40}{3}\right) \) (intersection of \( 2x + 5y = 100 \) and \( 8x + 5y = 200 \))
- \( C(0, 20) \)
Let us evaluate \( Z \) at each corner point:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(25, 0) \): \( Z = 25 \)
- At \( B\left(\frac{50}{3}, \frac{40}{3}\right) \): \( Z = \frac{50}{3} + \frac{40}{3} = \frac{90}{3} = 30 \)
- At \( C(0, 20) \): \( Z = 20 \)
Comparing these values, the maximum distance is 30 km.
\( \implies \) The young man can travel a maximum distance of 30 km if he rides \( \frac{50}{3} \) km at 25 km/h and \( \frac{40}{3} \) km at 40 km/h.
In simple words: Define distance travelled at both speeds as \( x \) and \( y \). Establish inequalities for fuel budget and time (using distance divided by speed), then locate the point that maximizes the total distance.

Exam Tip: Express time constraints as \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \), and convert fractions by taking the LCM to simplify the resulting inequality before graphing.

Question 9. A producer has 20 and 10 units of labour and capital respectively which he can use to produce two kinds of goods X and Y. To produce one unit of X, 2 units of capital and 1 unit of labour is required. To produce one unit of Y, 3 units of labour and one unit of capital is required. If X and Y are priced at Rs. 80 and Rs. 100 per unit respectively, how should the producer use his resources to maximise the total revenue?
Answer: Let \( x \) units of goods X and \( y \) units of goods Y be produced.
The objective function to maximize total revenue is:
\( Z = 80x + 100y \)
Subject to the constraints:
- Labour limit: \( x + 3y \le 20 \)
- Capital limit: \( 2x + y \le 10 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us determine the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( A(5, 0) \) (from \( 2x = 10 \))
- \( B(2, 6) \) (intersection of \( x + 3y = 20 \) and \( 2x + y = 10 \))
- \( C(0, 6.67) \) (from \( 3y = 20 \))
Let us evaluate the revenue \( Z \) at each corner point:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(5, 0) \): \( Z = 80(5) + 100(0) = 400 \)
- At \( B(2, 6) \): \( Z = 80(2) + 100(6) = 160 + 600 = 760 \)
- At \( C(0, 20/3) \): \( Z = 80(0) + 100(20/3) = 666.67 \)
Comparing these values, the maximum revenue is Rs. 760.
\( \implies \) The producer should produce 2 units of goods X and 6 units of goods Y to maximize his total revenue.
In simple words: Represent the products as variables. Set up constraints for available labor and capital, find the corners of the region, and calculate the maximum revenue.

Exam Tip: Carefully map which resources (labor vs. capital) apply to which product to avoid setting up incorrect coefficient equations.

Question 10. A factory owner purchases two types of machines A and B for his factory. The requirements and limitations for the machines are as follows:
 

He has maximum area of 9000 m2 available and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximise the daily output.
Answer: Let the factory owner buy \( x \) machines of type A and \( y \) machines of type B.
The objective function to maximize daily output is:
\( Z = 60x + 40y \)
Subject to the constraints:
- Floor area constraint: \( 1000x + 1200y \le 9000 \implies 5x + 6y \le 45 \)
- Labour force constraint: \( 12x + 8y \le 72 \implies 3x + 2y \le 18 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us find the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( A(6, 0) \)
- \( B(2.25, 5.625) \) (intersection of \( 5x + 6y = 45 \) and \( 3x + 2y = 18 \))
- \( C(0, 7.5) \)
Since the number of machines purchased must be whole integers, we evaluate integer coordinates near \( B(2.25, 5.625) \):
- For \( (4, 3) \): Area \( = 4(1000) + 3(1200) = 7600 \le 9000 \), Men \( = 4(12) + 3(8) = 72 \le 72 \). Daily output \( Z = 60(4) + 40(3) = 360 \) units.
- For \( (6, 0) \): Area \( = 6000 \le 9000 \), Men \( = 72 \le 72 \). Daily output \( Z = 60(6) + 40(0) = 360 \) units.
- For \( (3, 4) \): Area \( = 7800 \le 9000 \), Men \( = 68 \le 72 \). Daily output \( Z = 340 \) units.
\( \implies \) To achieve the maximum daily output of 360 units, the owner should either buy 6 machines of type A and 0 of type B, or buy 4 machines of type A and 3 of type B.
In simple words: Set up floor area and labor constraints. Because you cannot purchase a fraction of a machine, check the output for integer combinations near the intersection point to find the maximum possible output.

Exam Tip: Whenever an intersection point consists of decimals, check the surrounding integer coordinates that satisfy all constraints to find the correct realistic solution.

Question 11. A manufacturer makes two types of cups A and B. Three machines are required to manufacture the cups and the time in minutes required by each in as given below :
 

Each machine is available for a maximum period of 6 hours per day. If the profit on each cup A is 75 paise and on B is 50 paise, find how many cups of each type should be manufactured to maximise the profit per day.
Answer: Let \( x \) cups of type A and \( y \) cups of type B be produced.
Each machine is available for at most 6 hours (equal to 360 minutes).
The objective function to maximize total profit is:
\( Z = 0.75x + 0.50y \) (in Rs.)
Subject to the constraints:
- Machine I time: \( 12x + 6y \le 360 \implies 2x + y \le 60 \)
- Machine II time: \( 18x \le 360 \implies x \le 20 \)
- Machine III time: \( 6x + 9y \le 360 \implies 2x + 3y \le 120 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
The corner points of the bounded feasible region are:
- \( O(0, 0) \)
- \( A(20, 0) \)
- \( B(20, 20) \) (intersection of \( x = 20 \) and \( 2x + y = 60 \))
- \( C(15, 30) \) (intersection of \( 2x + y = 60 \) and \( 2x + 3y = 120 \))
- \( D(0, 40) \) (intersection of \( 2x + 3y = 120 \) and \( x = 0 \))
Let us evaluate the profit \( Z \) at each corner point:
- At \( A(20, 0) \): \( Z = 0.75(20) + 0 = 15.00 \)
- At \( B(20, 20) \): \( Z = 0.75(20) + 0.50(20) = 15 + 10 = 25.00 \)
- At \( C(15, 30) \): \( Z = 0.75(15) + 0.50(30) = 11.25 + 15 = 26.25 \)
- At \( D(0, 40) \): \( Z = 0 + 0.50(40) = 20.00 \)
Comparing these values, the maximum profit is Rs. 26.25.
\( \implies \) The manufacturer should produce 15 cups of type A and 30 cups of type B daily to maximize profit.
In simple words: Convert the machine time limits from hours to minutes. Formulate constraints for all three machines, locate the corner points of the feasible region, and pick the production values that yield the maximum profit.

Exam Tip: Pay close attention to units. Always convert the machine time limits from hours (6 hours) to minutes (360 minutes) to match the times provided in the table.

Question 12. A company produces two types of belts A and B. Profits on these belts are Rs. 2 and Rs. 1.50 per belt respectively. A belt of type A requires twice as much time as belt of type B. The company can produce almost 1000 belts of type B per day. Material for 800 belts per day is available. Almost 400 buckles for belts of type A and 700 for type B are available per day. How much belts of each type should the company produce so as to maximize the profit?
Answer: Let \( x \) be the number of belts of type A produced, and \( y \) be the number of belts of type B produced.
The objective function to maximize total profit is:
\( Z = 2x + 1.5y \)
Subject to the constraints:
- Time constraint: \( 2x + y \le 1000 \) (since A requires twice as much time as B, and the maximum production capacity of B alone is 1000)
- Material constraint: \( x + y \le 800 \)
- Buckles for A: \( x \le 400 \)
- Buckles for B: \( y \le 700 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
Let us find the corner points of the bounded feasible region:
- \( O(0, 0) \)
- \( E(400, 0) \)
- \( H(400, 200) \) (intersection of \( x = 400 \) and \( x + y = 800 \))
- \( G(200, 600) \) (intersection of \( x + y = 800 \) and \( 2x + y = 1000 \))
- \( F(0, 700) \) (intersection of \( y = 700 \) and \( 2x + y = 1000 \) in the first quadrant)
Let us evaluate the profit \( Z \) at each corner point:
- At \( E(400, 0) \): \( Z = 2(400) = 800 \)
- At \( H(400, 200) \): \( Z = 2(400) + 1.5(200) = 800 + 300 = 1100 \)
- At \( G(200, 600) \): \( Z = 2(200) + 1.5(600) = 400 + 900 = 1300 \)
- At \( F(0, 700) \): \( Z = 2(0) + 1.5(700) = 1050 \)
Comparing these values, the maximum profit is Rs. 1300.
\( \implies \) The company should produce 200 belts of type A and 600 belts of type B to maximize its daily profit.
In simple words: Represent the belts as \( x \) and \( y \). Express the constraints for time, leather material, and buckles, then locate the corner points of the region to find the maximum possible profit.

Exam Tip: The time constraint is formulated as \( 2x + y \le 1000 \), representing the relative time required to produce both types of belts.

Question 13. Two Godowns X and Y have a grain storage capacity of 100 quintals and 50 quintals respectively. Their supply goes to three ration shop A, B and C whose requirements are 60, 50 and 40 quintals respectively. The cost of transportation per quintals from the godowns to the shops are given in following table :
 

How should the supplies be transported to ration shops from godowns to minimize the transportation cost?
Answer: Let godown X transport \( x \) quintals of grain to shop A and \( y \) quintals to shop B.
Since the storage capacity of X is 100 quintals, the amount of grain transported from X to shop C is \( 100 - (x + y) \).
The remaining requirements of the shops are supplied by godown Y. Thus:
- Supply from Y to shop A: \( 60 - x \)
- Supply from Y to shop B: \( 50 - y \)
- Supply from Y to shop C: \( 40 - (100 - x - y) = x + y - 60 \)
The constraints based on capacities and requirements are:
- \( x \le 60 \), \( y \le 50 \)
- \( x + y \le 100 \)
- \( x + y \ge 60 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
The total transportation cost \( Z \) is:
\( Z = 6x + 3y + 2.5(100 - x - y) + 4(60 - x) + 2(50 - y) + 3(x + y - 60) \)
\( \implies Z = 2.5x + 1.5y + 410 \)
The corner points of the bounded feasible region are:
- \( P(10, 50) \)
- \( Q(50, 50) \)
- \( R(60, 40) \)
- \( S(60, 0) \)
Let us evaluate the total cost \( Z \) at each corner point:
- At \( P(10, 50) \): \( Z = 2.5(10) + 1.5(50) + 410 = 510 \)
- At \( Q(50, 50) \): \( Z = 2.5(50) + 1.5(50) + 410 = 610 \)
- At \( R(60, 40) \): \( Z = 2.5(60) + 1.5(40) + 410 = 620 \)
- At \( S(60, 0) \): \( Z = 2.5(60) + 1.5(0) + 410 = 560 \)
Comparing these values, the minimum cost is Rs. 510.
\( \implies \) For minimum transportation cost of Rs. 510:
- Godown X transports 10 quintals to A, 50 quintals to B, and 40 quintals to C.
- Godown Y transports 50 quintals to A, 0 quintals to B, and 0 quintals to C.
In simple words: Represent the quantities shipped from X to A and B as variables, expressing all other shipments in terms of these two variables. Set up constraints to ensure capacities and demands are respected, and then find the combination that minimizes the cost equation.

Exam Tip: In transportation problems, defining the variables \( x \) and \( y \) as the shipments from a single source helps simplify the problem into two variables instead of six.

Question 14. An Aeroplane can carry a 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 atleast four times as many passengers prefer to travel by second class than by first class. Determine, how many tickets of each type must be sold to maximize profit for the airline.
Answer: Let \( x \) be the number of first class tickets sold, and \( y \) be the number of second class tickets sold.
The objective function to maximize total profit is:
\( Z = 400x + 300y \)
Subject to the constraints:
- Total passengers: \( x + y \le 200 \)
- Seat reservation: \( x \ge 20 \)
- Seat preference: \( y \ge 4x \implies 4x - y \le 0 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
The corner points of the bounded feasible region are:
- \( A(20, 80) \) (intersection of \( x = 20 \) and \( y = 4x \) )
- \( B(40, 160) \) (intersection of \( x + y = 200 \) and \( y = 4x \) )
- \( C(20, 180) \) (intersection of \( x = 20 \) and \( x + y = 200 \) )
Let us evaluate the profit \( Z \) at each corner point:
- At \( A(20, 80) \): \( Z = 400(20) + 300(80) = 8000 + 24000 = 32000 \)
- At \( B(40, 160) \): \( Z = 400(40) + 300(160) = 16000 + 48000 = 64000 \)
- At \( C(20, 180) \): \( Z = 400(20) + 300(180) = 8000 + 54000 = 62000 \)
Comparing these values, the maximum profit is Rs. 64,000.
\( \implies \) The airline must sell 40 first class tickets and 160 second class tickets to obtain the maximum profit of Rs. 64,000.
In simple words: Let \( x \) be first class and \( y \) be second class tickets. Formulate constraints for the total seats, reserved seats, and passenger preferences. Evaluate the profit at all corner points to find the best configuration.

Exam Tip: Pay close attention to the phrasing "at least four times as many prefer second class to first class". This translates to \( y \ge 4x \), which is a common source of confusion.

Question 15. A diet for a sick person must contain atleast 4000 units of vitamins, 50 units of minerals and 1400 units of calories. Two foods A and B are available at a cost of Rs. 5 and Rs. 4 per unit respectively. One unit of food A contains 200 unit of vitamins, 1 unit of minerals and 40 units of calories whereas one unit of food B contains 100 units of vitamins, 2 units of minerals and 40 units of calories. Find what combination of the food A and B should be used to have least cost but it must satisfy the requirements of the sick person.
Answer: Let \( x \) units of food A and \( y \) units of food B be used in the diet.
The objective function to minimize total cost is:
\( Z = 5x + 4y \)
Subject to the constraints:
- Vitamins requirement: \( 200x + 100y \ge 4000 \implies 2x + y \ge 40 \)
- Minerals requirement: \( x + 2y \ge 50 \)
- Calories requirement: \( 40x + 40y \ge 1400 \implies x + y \ge 35 \)
- Non-negativity: \( x \ge 0, y \ge 0 \)
The corner points of the unbounded feasible region are:
- \( A(0, 40) \)
- \( B(5, 30) \) (intersection of \( 2x + y = 40 \) and \( x + y = 35 \) )
- \( C(20, 15) \) (intersection of \( x + y = 35 \) and \( x + 2y = 50 \) )
- \( D(50, 0) \)
Let us evaluate the total cost \( Z \) at each corner point:
- At \( A(0, 40) \): \( Z = 5(0) + 4(40) = 160 \)
- At \( B(5, 30) \): \( Z = 5(5) + 4(30) = 25 + 120 = 145 \)
- At \( C(20, 15) \): \( Z = 5(20) + 4(15) = 100 + 60 = 160 \)
- At \( D(50, 0) \): \( Z = 5(50) + 4(0) = 250 \)
Since the feasible region is unbounded, we verify the minimum value by plotting the inequality \( 5x + 4y < 145 \). Since this open half-plane has no common points with our feasible region, the minimum is confirmed.
\( \implies \) The least cost combination is 5 units of food A and 30 units of food B, yielding a minimum cost of Rs. 145.
In simple words: Represent the quantities of the two foods as \( x \) and \( y \). Write equations based on the minimum requirements for vitamins, minerals, and calories. Solve at the corners to find the cheapest food combination.

Exam Tip: In diet optimization problems, ensure you represent "at least" requirements using "greater-than-or-equal-to" (\( \ge \)) signs, and clearly show the verification of the minimum value for unbounded regions.