CBSE Class 12 Mathematics Linear Programming Assignment Set 01

Some Important Results/Concepts

Solving linear programming problem using Corner Point Method. The method comprises of the following steps:

1. Find the feasible region of the linear programming problem and determine its corner points (vertices) either by inspection or by solving the two equations of the lines intersecting at that point.

2. Evaluate the objective function \( Z = ax + by \) at each corner point. Let \( M \) and \( m \), respectively denote the largest and smallest values of these points.

3. (i) When the feasible region is bounded, \( M \) and \( m \) are the maximum and minimum values of \( Z \).
(ii) In case the feasible region is unbounded, we have:

4. (a) \( M \) is the maximum value of \( Z \), if the open half plane determined by \( ax + by > M \) has no point in common with the feasible region. Otherwise, \( Z \) has no maximum value.
(b) Similarly, \( m \) is the minimum value of \( Z \), if the open half plane determined by \( ax + by < m \) has no point in common with the feasible region. Otherwise, \( Z \) has no minimum value.

Assignments

(i) LPP and its Mathematical Formulation

Level I

Question 1. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain atleast 8 units of vitamin A and 10 units of vitamin C. Food ‘I’ contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food ‘II’ contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem.
Answer:
Let \( x \) kg be the quantity of Food I and \( y \) kg be the quantity of Food II in the mixture. Since the quantity of foods cannot be negative, we have:
\( x \ge 0, \quad y \ge 0 \)
The total cost of purchasing these foods is given by \( Z = 50x + 70y \), which is our objective function to be minimized.
Now, we set up the constraints based on the minimum requirements of vitamins:
For Vitamin A, the mixture must have at least 8 units:
\( 2x + y \ge 8 \)
For Vitamin C, the mixture must have at least 10 units:
\( x + 2y \ge 10 \)
Hence, the linear programming problem is formulated as:
Minimize \( Z = 50x + 70y \)
subject to the constraints:
\( 2x + y \ge 8 \)
\( x + 2y \ge 10 \)
\( x, y \ge 0 \)
In simple words: Let \( x \) and \( y \) be the weights of Food I and Food II. We want to spend the least amount of money, so we minimize the cost equation \( 50x + 70y \) while ensuring we meet the minimum limits of 8 units for Vitamin A and 10 units for Vitamin C.

Exam Tip: Always state the non-negativity constraints \( x \ge 0, y \ge 0 \) explicitly, as they carry marks in formulation questions.

(ii) Graphical Method of Solving LPP (Bounded and Unbounded Solutions)

Level I

Question 1. Minimise Z = – 3x + 4 ysubject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.
Answer:
We first convert the given inequalities into equations to plot the boundary lines:
1. \( x + 2y = 8 \)
- Passes through \( (8, 0) \) and \( (0, 4) \).
2. \( 3x + 2y = 12 \)
- Passes through \( (4, 0) \) and \( (0, 6) \).
The intersection point of these two lines is found by solving them simultaneously:
Subtracting the first equation from the second gives:
\( 2x = 4 \)

\(\implies x = 2 \)
Substituting \( x = 2 \) into \( x + 2y = 8 \) gives \( y = 3 \). Thus, the lines intersect at \( (2, 3) \).
The feasible region is bounded by \( x \ge 0 \), \( y \ge 0 \), and the two lines. The vertices (corner points) of this region are:
- \( O(0, 0) \)
- \( A(4, 0) \)
- \( B(2, 3) \)
- \( C(0, 4) \)
Let us evaluate the objective function \( Z = -3x + 4y \) at each vertex:
- At \( O(0, 0) \): \( Z = -3(0) + 4(0) = 0 \)
- At \( A(4, 0) \): \( Z = -3(4) + 4(0) = -12 \)
- At \( B(2, 3) \): \( Z = -3(2) + 4(3) = 6 \)
- At \( C(0, 4) \): \( Z = -3(0) + 4(4) = 16 \)
Comparing these values, the minimum value of \( Z \) is \( -12 \), which occurs at the point \( (4, 0) \).
In simple words: Draw the boundary lines of the inequalities on a graph to find the shaded area where all conditions are met. Check the value of \( Z \) at each corner of this area. The lowest value is \( -12 \) at the corner \( (4, 0) \).

Exam Tip: Clearly show the calculation for the intersection point of the boundary lines to secure full marks for graph-based questions.

Question 2. Maximise Z = 5x + 3ysubject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.
Answer:
We convert the given inequalities into boundary line equations:
1. \( 3x + 5y = 15 \)
- Passes through \( (5, 0) \) and \( (0, 3) \).
2. \( 5x + 2y = 10 \)
- Passes through \( (2, 0) \) and \( (0, 5) \).
To find the intersection of these two lines, we solve them simultaneously:
Multiply the first equation by 2 and the second by 5:
\( 6x + 10y = 30 \)
\( 25x + 10y = 50 \)
Subtracting the first equation from the second:
\( 19x = 20 \)

\(\implies x = \frac{20}{19} \)
Substituting \( x = \frac{20}{19} \) back into the second equation gives \( y = \frac{45}{19} \).
The intersection point is \( \left(\frac{20}{19}, \frac{45}{19}\right) \).
The corner points of the bounded feasible region are:
- \( O(0, 0) \)
- \( A(2, 0) \)
- \( B\left(\frac{20}{19}, \frac{45}{19}\right) \)
- \( C(0, 3) \)
Let us evaluate \( Z = 5x + 3y \) at each of these corner points:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(2, 0) \): \( Z = 5(2) + 3(0) = 10 \)
- At \( B\left(\frac{20}{19}, \frac{45}{19}\right) \): \( Z = 5\left(\frac{20}{19}\right) + 3\left(\frac{45}{19}\right) = \frac{235}{19} \approx 12.37 \)
- At \( C(0, 3) \): \( Z = 5(0) + 3(3) = 9 \)
Comparing these values, the maximum value of \( Z \) is \( \frac{235}{19} \) (or approximately 12.37), which is achieved at the point \( \left(\frac{20}{19}, \frac{45}{19}\right) \).
In simple words: Plot the lines on a graph. Find the common shaded area, then test \( Z \) at each corner of this shape. The highest value is about 12.37, which is found at the corner where the two lines cross.

Exam Tip: Keep your corner points as exact fractions rather than rounding them to decimals to ensure you don't lose any points for accuracy.

Question 3. Minimise Z = 3x + 5y suchthat x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.
Answer:
We first find the boundary lines by converting the inequalities into equations:
1. \( x + 3y = 3 \)
- Passes through \( (3, 0) \) and \( (0, 1) \).
2. \( x + y = 2 \)
- Passes through \( (2, 0) \) and \( (0, 2) \).
The intersection point of these two lines is found by solving them simultaneously:
Subtracting the second equation from the first:
\( 2y = 1 \)

\(\implies y = \frac{1}{2} \)
Substituting \( y = \frac{1}{2} \) back into \( x + y = 2 \) gives \( x = \frac{3}{2} \).
So, the intersection point is \( \left(\frac{3}{2}, \frac{1}{2}\right) \).
Since the constraints are of the type \( \ge \), the feasible region is unbounded and lies away from the origin. The corner points of this region are:
- \( A(3, 0) \)
- \( B\left(\frac{3}{2}, \frac{1}{2}\right) \)
- \( C(0, 2) \)
Let us find the value of \( Z = 3x + 5y \) at these vertices:
- At \( A(3, 0) \): \( Z = 3(3) + 5(0) = 9 \)
- At \( B\left(\frac{3}{2}, \frac{1}{2}\right) \): \( Z = 3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = 7 \)
- At \( C(0, 2) \): \( Z = 3(0) + 5(2) = 10 \)
Since the region is unbounded, we must check if the open half-plane \( 3x + 5y < 7 \) has any point in common with the feasible region.
Since there is no common point between the open half-plane \( 3x + 5y < 7 \) and the feasible region, the minimum value of \( Z \) is \( 7 \) at the point \( \left(\frac{3}{2}, \frac{1}{2}\right) \).
In simple words: This problem has an open-ended shaded region going outwards. The lowest value we calculate at the corners is 7. Since the region \( 3x + 5y < 7 \) doesn't overlap with our shaded area, 7 is indeed the absolute minimum.

Exam Tip: For unbounded regions, you must always perform the half-plane test (plotting the dotted line of the objective function) to prove whether the minimum or maximum exists.

(iii) Diet Problem

Level II

Question 1. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1,400 calories. Two foods X and Y are available at a cost of Rs. 4 and Rs. 3 per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of mineral and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of X and Y should be used to have least cost? Also find the least cost.
Answer:
Let \( x \) units of food X and \( y \) units of food Y be included in the diet. Since quantities cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to minimize is the total cost:
Minimize \( Z = 4x + 3y \)
Subject to the constraints:
1. Vitamins requirement (at least 4000 units):
\( 200x + 100y \ge 4000 \implies 2x + y \ge 40 \)
2. Minerals requirement (at least 50 units):
\( x + 2y \ge 50 \)
3. Calories requirement (at least 1400 calories):
\( 40x + 40y \ge 1400 \implies x + y \ge 35 \)
Let us plot the lines to find the corner points of the unbounded feasible region:
- The line \( 2x + y = 40 \) passes through \( (20, 0) \) and \( (0, 40) \).
- The line \( x + 2y = 50 \) passes through \( (50, 0) \) and \( (0, 25) \).
- The line \( x + y = 35 \) passes through \( (35, 0) \) and \( (0, 35) \).
By solving the equations of intersecting lines, we find the corner points of the feasible region:
- \( A(50, 0) \)
- \( B(20, 15) \) [intersection of \( x + 2y = 50 \) and \( x + y = 35 \)]
- \( C(5, 30) \) [intersection of \( 2x + y = 40 \) and \( x + y = 35 \)]
- \( D(0, 40) \)
Now, let us find the value of \( Z = 4x + 3y \) at each corner point:
- At \( A(50, 0) \): \( Z = 4(50) + 3(0) = 200 \)
- At \( B(20, 15) \): \( Z = 4(20) + 3(15) = 125 \)
- At \( C(5, 30) \): \( Z = 4(5) + 3(30) = 110 \)
- At \( D(0, 40) \): \( Z = 4(0) + 3(40) = 120 \)
Since the region is unbounded, we verify that the open half-plane \( 4x + 3y < 110 \) shares no points with the feasible region.
Therefore, the least cost is Rs. 110, which is obtained by using 5 units of food X and 30 units of food Y.
In simple words: To find the cheapest way to meet all health targets, we solve the equations on a graph. Testing the corners of the safe region shows that 5 units of Food X and 30 units of Food Y cost the least at Rs. 110.

Exam Tip: Be sure to write down the equations and how you solved them to get the coordinates of the corner points like \( (5, 30) \).

Question 2. Every gram of wheat provides 0.1 g of proteins and 0.25 g of carbohydrates. The corresponding values for rice are 0.05 g and 0.5 g respectively. Wheat costs Rs. 10 per kg and rice Rs. 20 per kg. The minimum daily requirements of protein and carbohydrates for an average child are 50 gm and 200 gm respectively. In what quantities, should wheat and rice be mixed in the daily diet to provide the minimum daily requirements of protein and carbohydrates at minimum cost ?
Answer:
Let \( x \) kg of wheat and \( y \) kg of rice be mixed in the daily diet. Since the quantities cannot be negative, we have \( x \ge 0, y \ge 0 \).
Since 1 kg of wheat provides \( 0.1 \times 1000 = 100 \) g of protein and \( 0.25 \times 1000 = 250 \) g of carbohydrates, and 1 kg of rice provides \( 0.05 \times 1000 = 50 \) g of protein and \( 0.5 \times 1000 = 500 \) g of carbohydrates:
The LPP is formulated as:
Minimize \( Z = 10x + 20y \)
Subject to the constraints:
1. Protein: \( 100x + 50y \ge 50 \implies 2x + y \ge 1 \)
2. Carbohydrates: \( 250x + 500y \ge 200 \implies x + 2y \ge 0.8 \)
Let us find the corner points of the unbounded feasible region:
- The line \( 2x + y = 1 \) passes through \( (0.5, 0) \) and \( (0, 1) \).
- The line \( x + 2y = 0.8 \) passes through \( (0.8, 0) \) and \( (0, 0.4) \).
The intersection of these two lines is at \( (0.4, 0.2) \).
The corner points of the feasible region are:
- \( A(0.8, 0) \)
- \( B(0.4, 0.2) \)
- \( C(0, 1) \)
Let us evaluate \( Z = 10x + 20y \) at these points:
- At \( A(0.8, 0) \): \( Z = 10(0.8) + 20(0) = 8 \)
- At \( B(0.4, 0.2) \): \( Z = 10(0.4) + 20(0.2) = 8 \)
- At \( C(0, 1) \): \( Z = 10(0) + 20(1) = 20 \)
The minimum cost of Rs. 8 occurs at all points along the line segment joining \( A(0.8, 0) \) and \( B(0.4, 0.2) \). For a mix containing both wheat and rice, we choose \( 0.4 \) kg (or 400 gm) of wheat and \( 0.2 \) kg (or 200 gm) of rice.
In simple words: To keep costs lowest, mix 400 gm of wheat and 200 gm of rice. This combination fully satisfies the child's minimum protein and carb needs for just Rs. 8.

Exam Tip: If two corner points yield the same minimum value, then any point on the line segment connecting them is also an optimal solution.

(iv) Manufacturing Problem

Level II

Question 1. A company manufactures two articles A and B. There are two departments through which these articles are processed: (i) assembly and (ii) finishing departments. The maximum capacity of the assembly department is 60 hours a week and that of the finishing department is 48 hours a week. The production of each article A requires 4 hours in assembly and 2 hours in finishing and that of each unit of B requires 2 hours in assembly and 4 hours in finishing. If the profit is Rs. 6 for each unit of A and Rs. 8 for each unit of B, find the number of units of A and B to be produced per week in order to have maximum profit.
Answer:
Let \( x \) be the number of units of article A and \( y \) be the number of units of article B produced per week. Since production cannot be negative, we have \( x \ge 0, y \ge 0 \).
The objective function to maximize is the total profit:
Maximize \( Z = 6x + 8y \)
Subject to the capacity constraints:
1. Assembly department constraint (maximum 60 hours):
\( 4x + 2y \le 60 \implies 2x + y \le 30 \)
2. Finishing department constraint (maximum 48 hours):
\( 2x + 4y \le 48 \implies x + 2y \le 24 \)
Let us plot the equations of the boundary lines:
- The line \( 2x + y = 30 \) has intercepts \( (15, 0) \) and \( (0, 30) \).
- The line \( x + 2y = 24 \) has intercepts \( (24, 0) \) and \( (0, 12) \).
Solving the two equations simultaneously gives their intersection point:
\( 2x + y = 30 \implies y = 30 - 2x \)
Substituting into \( x + 2y = 24 \):
\( x + 2(30 - 2x) = 24 \implies 3x = 36 \implies x = 12 \)
Then, \( y = 30 - 2(12) = 6 \). So the intersection is at \( (12, 6) \).
The corner points of the bounded feasible region are:
- \( O(0, 0) \)
- \( A(15, 0) \)
- \( B(12, 6) \)
- \( C(0, 12) \)
Evaluating the profit \( Z = 6x + 8y \) at each vertex:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(15, 0) \): \( Z = 6(15) = 90 \)
- At \( B(12, 6) \): \( Z = 6(12) + 8(6) = 120 \)
- At \( C(0, 12) \): \( Z = 8(12) = 96 \)
Thus, the maximum profit of Rs. 120 is obtained when 12 units of article A and 6 units of article B are manufactured per week.
In simple words: To maximize profit, the company should make 12 units of product A and 6 units of product B each week. This keeps the factory fully utilized within its weekly limits and brings in a peak profit of Rs. 120.

Exam Tip: Be sure to draw/mention the shaded feasible region clearly to show how the corner points are selected.

Question 2. A company sells two different produces A and B. The two products are produced in a common production process which has a total capacity of 500 man hours. It takes 5 hours to produce a unit of A and 3 hours to produce a unit of B. The demand in the market shows that the maximum number of units of A that can be sold is 70 and that for B is 125. Profit on each unit of A is Rs. 20 and that on B is Rs. 15. How many units of A and B should be produced to maximize the profit? Solve it graphically
Answer:
Let \( x \) and \( y \) be the number of units of products A and B produced respectively. Since quantities cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to maximize is the total profit:
Maximize \( Z = 20x + 15y \)
Subject to the constraints:
1. Production capacity: \( 5x + 3y \le 500 \)
2. Market demand for A: \( x \le 70 \)
3. Market demand for B: \( y \le 125 \)
Let us find the vertices (corner points) of the bounded feasible region:
- Point \( O(0, 0) \)
- Point \( A(70, 0) \)
- Point \( B(70, 50) \) [intersection of \( x = 70 \) and \( 5x + 3y = 500 \)]
- Point \( C(25, 125) \) [intersection of \( y = 125 \) and \( 5x + 3y = 500 \)]
- Point \( D(0, 125) \)
Evaluating the profit function \( Z = 20x + 15y \) at each vertex:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(70, 0) \): \( Z = 20(70) + 15(0) = 1400 \)
- At \( B(70, 50) \): \( Z = 20(70) + 15(50) = 2150 \)
- At \( C(25, 125) \): \( Z = 20(25) + 15(125) = 2375 \)
- At \( D(0, 125) \): \( Z = 20(0) + 15(125) = 1875 \)
Thus, the maximum profit of Rs. 2375 is obtained when the company produces 25 units of product A and 125 units of product B.
In simple words: Test the corner points on the graph. The highest profit is Rs. 2375, which we get by making 25 units of product A and 125 units of product B.

Exam Tip: Always make sure to verify that the intersection points lie within the individual limit constraints (e.g., \( y \le 125 \) and \( x \le 70 \)).

Level III

Question 1. A manufacture makes two types of cups, A and B. Three machines are required to manufacture the cups and the time in minutes required by each is 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 it is 50 paise, show that the 15 cups of type A and 30 cups of type B should be manufactured per day to get the maximum profit.
Answer:
Let the manufacturer produce \( x \) cups of type A and \( y \) cups of type B per day. Since production counts cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The total profit in Rupees is given by:
Maximize \( Z = 0.75x + 0.50y \)
Since each machine is available for at most 6 hours (i.e., \( 6 \times 60 = 360 \) minutes) per day, the constraints are:
1. Machine I: \( 12x + 6y \le 360 \implies 2x + y \le 60 \)
2. Machine II: \( 18x \le 360 \implies x \le 20 \)
3. Machine III: \( 6x + 9y \le 360 \implies 2x + 3y \le 120 \)
Let us determine the vertices of the bounded feasible region by analyzing the boundary line equations:
- The line \( 2x + y = 60 \) has intercepts \( (30, 0) \) and \( (0, 60) \).
- The line \( x = 20 \) is vertical.
- The line \( 2x + 3y = 120 \) has intercepts \( (60, 0) \) and \( (0, 40) \).
The corner points of this 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 \( 2x + 3y = 120 \)]
- \( D(0, 40) \)
Let us evaluate \( Z = 0.75x + 0.50y \) at these vertices:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(20, 0) \): \( Z = 0.75(20) = 15 \)
- At \( B(20, 20) \): \( Z = 0.75(20) + 0.50(20) = 25 \)
- At \( C(15, 30) \): \( Z = 0.75(15) + 0.50(30) = 26.25 \)
- At \( D(0, 40) \): \( Z = 0.50(40) = 20 \)
Comparing these values, the maximum profit is Rs. 26.25, which occurs when 15 cups of type A and 30 cups of type B are produced daily.
In simple words: Converting the machine hours to minutes gives us three boundary inequalities. Checking the corner points of the bounded area on the graph confirms that making 15 cups of type A and 30 cups of type B yields the maximum possible daily profit of Rs. 26.25.

Exam Tip: Be sure to convert hours to minutes to keep all units consistent across your constraint inequalities.

(v) Allocation Problem

Level II

Question 1. Ramesh wants to invest at most Rs. 70,000 in Bonds A and B. According to the rules, he has to invest at least Rs. 10,000 in Bond A and at least Rs. 30,000 in Bond B. If the rate of interest on bond A is 8 % per annum and the rate of interest on bond B is 10 % per annum , how much money should he invest to earn maximum yearly income ? Find also his maximum yearly income.
Answer:
Let Rs. \( x \) and Rs. \( y \) be invested in Bond A and Bond B respectively. The non-negativity constraints are \( x \ge 0, y \ge 0 \).
The objective function to maximize is the yearly income:
Maximize \( Z = 0.08x + 0.10y \)
Subject to the constraints:
1. Total investment limit: \( x + y \le 70000 \)
2. Minimum investment in Bond A: \( x \ge 10000 \)
3. Minimum investment in Bond B: \( y \ge 30000 \)
The boundary lines are:
- \( x + y = 70000 \)
- \( x = 10000 \)
- \( y = 30000 \)
The corner points of the bounded triangular feasible region are:
- \( P(10000, 30000) \)
- \( Q(40000, 30000) \)
- \( R(10000, 60000) \)
Evaluating the objective function at these vertices:
- At \( P(10000, 30000) \): \( Z = 0.08(10000) + 0.10(30000) = 800 + 3000 = 3800 \)
- At \( Q(40000, 30000) \): \( Z = 0.08(40000) + 0.10(30000) = 3200 + 3000 = 6200 \)
- At \( R(10000, 60000) \): \( Z = 0.08(10000) + 0.10(60000) = 800 + 6000 = 6800 \)
Thus, to maximize his yearly income, Ramesh should invest Rs. 10,000 in Bond A and Rs. 60,000 in Bond B. His maximum annual income will be Rs. 6,800.
In simple words: Since Bond B pays a higher interest rate (10% vs 8%), Ramesh should put as much money into Bond B as possible after meeting the minimum requirement of Rs. 10,000 for Bond A. This means investing Rs. 10,000 in Bond A and Rs. 60,000 in Bond B to earn Rs. 6,800.

Exam Tip: Since this is a simple linear optimization, you can quickly verify your answer: the higher-yielding bond should always get the maximum allocation possible.

Question 2. An oil company requires 12,000, 20,000 and 15,000 barrels of high grade, medium grade and low grade oil respectively. Refinery A produces 100, 300 and 200 barrels per day of high, medium and low grade oil respectively whereas the Refinery B produces 200, 400 and 100 barrels per day respectively. If A costs Rs. 400 per day and B costs Rs. 300 per day to operate, how many days should each be run to minimize the cost of requirement?
Answer:
Let Refinery A be run for \( x \) days and Refinery B for \( y \) days. Since the number of operating days cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to minimize is the total cost:
Minimize \( Z = 400x + 300y \)
Subject to the grade requirements:
1. High grade requirement: \( 100x + 200y \ge 12000 \implies x + 2y \ge 120 \)
2. Medium grade requirement: \( 300x + 400y \ge 20000 \implies 3x + 4y \ge 200 \)
3. Low grade requirement: \( 200x + 100y \ge 15000 \implies 2x + y \ge 150 \)
Let us identify the boundary vertices of the unbounded feasible region:
- Point \( A(120, 0) \)
- Point \( B(60, 30) \) [intersection of \( x + 2y = 120 \) and \( 2x + y = 150 \)]
- Point \( C(0, 150) \)
Now, let us evaluate the objective function at these vertices:
- At \( A(120, 0) \): \( Z = 400(120) + 300(0) = 48000 \)
- At \( B(60, 30) \): \( Z = 400(60) + 300(30) = 33000 \)
- At \( C(0, 150) \): \( Z = 400(0) + 300(150) = 45000 \)
Comparing the costs, the minimum operating cost is Rs. 33,000, which occurs when Refinery A runs for 60 days and Refinery B runs for 30 days.
In simple words: To meet the demand for all three oil grades at the lowest expense, the company should run Refinery A for 60 days and Refinery B for 30 days. This costs Rs. 33,000.

Exam Tip: Be sure to simplify constraints by dividing by common factors (like dividing by 100) before solving to keep the calculations simple.

Level III

Question 1. An aeroplane can carry a maximum of 250 passengers. A profit of Rs 500 is made on each executive class ticket and a profit of Rs 350 is made on each economy class ticket. The airline reserves at least 25 seats for executive class. However, at least 3 times as many passengers prefer to travel by economy class than by the executive class. Determine how many tickets of each type must be sold in order to maximize the profit for the airline. What is the maximum profit?
Answer:
Let \( x \) and \( y \) be the number of executive class and economy class tickets sold respectively. Since ticket quantities cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to maximize is the total profit:
Maximize \( Z = 500x + 350y \)
Subject to the constraints:
1. Maximum capacity constraint: \( x + y \le 250 \)
2. Minimum executive seats: \( x \ge 25 \)
3. Passenger preference: \( y \ge 3x \implies 3x - y \le 0 \)
The boundary lines are:
- \( x + y = 250 \)
- \( x = 25 \)
- \( y = 3x \)
The corner points of the bounded feasible region are:
- \( P(25, 75) \) [intersection of \( x = 25 \) and \( y = 3x \)]
- \( Q(25, 225) \) [intersection of \( x = 25 \) and \( x + y = 250 \)]
- \( R(62.5, 187.5) \) [intersection of \( y = 3x \) and \( x + y = 250 \)]
Evaluating the profit function at these corner points:
- At \( P(25, 75) \): \( Z = 500(25) + 350(75) = 38750 \)
- At \( Q(25, 225) \): \( Z = 500(25) + 350(225) = 91250 \)
- At \( R(62.5, 187.5) \): \( Z = 500(62.5) + 350(187.5) = 96875 \)
Since the number of tickets must be whole numbers, the nearest integer point within the feasible region is \( (62, 188) \).
- At \( (62, 188) \): \( Z = 500(62) + 350(188) = 96800 \)
Thus, the airline should sell 62 executive class tickets and 188 economy class tickets to obtain a maximum profit of Rs. 96,800.
In simple words: The mathematical corner is 62.5 and 187.5. Since you cannot sell half a ticket, we round to the nearest whole numbers inside the allowed area, which is 62 executive and 188 economy tickets, giving a maximum profit of Rs. 96,800.

Exam Tip: Always mention both the theoretical fractional corner point \( (62.5, 187.5) \) and the practical integer solution \( (62, 188) \) to avoid any penalty from examiners.

(vi) Transportation Problem

Level III

Question 1. A medicine company has factories at two places A and B . From these places, supply is to be made to each of its three agencies P, Q and R. The monthly requirement of these agencies are respectively 40, 40 and 50 packets of the medicines, While the production capacity of the factories at A and B are 60 and 70 packets are respectively. The transportation cost per packet from these factories to the agencies are given:

How many packets from each factory be transported to each agency so that the cost of transportation is minimum ? Also find the minimum cost.
Answer:
Let factory A deliver \( x \) packets to agency P and \( y \) packets to agency Q. Since A's total capacity is 60 packets, it delivers the remaining \( 60 - x - y \) packets to agency R.
The monthly requirements are 40 for P, 40 for Q, and 50 for R. Therefore, the quantities transported from B must be:
- To P: \( 40 - x \)
- To Q: \( 40 - y \)
- To R: \( 50 - (60 - x - y) = x + y - 10 \)
Since all transportation quantities must be non-negative:
\( x \ge 0, \quad y \ge 0, \quad x + y \le 60, \quad x \le 40, \quad y \le 40, \quad x + y \ge 10 \)
The objective function to minimize is the total cost \( Z \):
\( Z = 5x + 4y + 3(60 - x - y) + 4(40 - x) + 2(40 - y) + 5(x + y - 10) \)
\( Z = 3x + 4y + 370 \)
The corner points of the bounded feasible region are:
- \( A(10, 0) \)
- \( B(40, 0) \)
- \( C(40, 20) \)
- \( D(20, 40) \)
- \( E(0, 40) \)
- \( F(0, 10) \)
Evaluating the cost \( Z = 3x + 4y + 370 \) at each vertex:
- At \( A(10, 0) \): \( Z = 3(10) + 4(0) + 370 = 400 \)
- At \( B(40, 0) \): \( Z = 3(40) + 4(0) + 370 = 490 \)
- At \( C(40, 20) \): \( Z = 3(40) + 4(20) + 370 = 570 \)
- At \( D(20, 40) \): \( Z = 3(20) + 4(40) + 370 = 590 \)
- At \( E(0, 40) \): \( Z = 3(0) + 4(40) + 370 = 530 \)
- At \( F(0, 10) \): \( Z = 3(0) + 4(10) + 370 = 410 \)
Comparing these, the minimum cost is Rs. 400 at \( (10, 0) \).
The optimal distribution scheme is:
- From Factory A: 10 packets to P, 0 packets to Q, 50 packets to R.
- From Factory B: 30 packets to P, 40 packets to Q, 0 packets to R.
The minimum transportation cost is Rs. 400.
In simple words: By assigning helper variables \( x \) and \( y \) to represent shipments from Factory A, we formulate the cost equation \( 3x + 4y + 370 \). Evaluating the vertices of the safe region shows that Factory A should send 10 packets to P and 50 packets to R, while Factory B sends 30 packets to P and 40 packets to Q for a minimum cost of Rs. 400.

Exam Tip: Be sure to write the full set of non-negativity inequalities carefully, as each shipping route must be \( \ge 0 \).

Questions for Self Evaluation

Question 1. Solve the following linear programming problem graphically : Maximize z = x – 7y +190 subject to the constraints x + y  8, x  5, y  5, x + y  4, x  0, y  0 .
Answer:
Let us identify the vertices (corner points) of the bounded feasible region defined by the constraints:
- \( x + y \le 8 \)
- \( x \le 5 \)
- \( y \le 5 \)
- \( x + y \ge 4 \)
- \( x \ge 0, \quad y \ge 0 \)
The boundary lines are:
- \( x + y = 8 \)
- \( x = 5 \)
- \( y = 5 \)
- \( x + y = 4 \)
The corner points are:
- \( A(4, 0) \)
- \( B(5, 0) \)
- \( C(5, 3) \)
- \( D(3, 5) \)
- \( E(0, 5) \)
- \( F(0, 4) \)
Evaluating the objective function \( Z = x - 7y + 190 \) at each corner point:
- At \( A(4, 0) \): \( Z = 4 - 7(0) + 190 = 194 \)
- At \( B(5, 0) \): \( Z = 5 - 7(0) + 190 = 195 \)
- At \( C(5, 3) \): \( Z = 5 - 7(3) + 190 = 174 \)
- At \( D(3, 5) \): \( Z = 3 - 7(5) + 190 = 158 \)
- At \( E(0, 5) \): \( Z = 0 - 7(5) + 190 = 155 \)
- At \( F(0, 4) \): \( Z = 0 - 7(4) + 190 = 162 \)
Comparing these values, the maximum value of \( Z \) is \( 195 \) at the point \( (5, 0) \).
In simple words: Plot all constraint lines on a graph. Find the common shaded area, then check the value of \( Z \) at each corner. The highest value is 195, which is located at the corner point \( (5, 0) \).

Exam Tip: Be careful to read the signs in the inequalities. Draw a quick arrow on each boundary line indicating which side of the line is favored (above or below) to identify the correct feasible region.

Question 2. Solve the following linear programming problem graphically : Maximize z = 3x + 5y subject to the constraints x + y  2, x + 3y  3, x  0, y  0 .
Answer:
The boundary lines are:
1. \( x + y = 2 \)
2. \( x + 3y = 3 \)
Since both constraints are of the type \( \ge \), the feasible region is unbounded and extends infinitely in the first quadrant away from the origin.
The corner points of this unbounded region are:
- \( A(3, 0) \)
- \( B\left(\frac{3}{2}, \frac{1}{2}\right) \)
- \( C(0, 2) \)
Let us evaluate \( Z = 3x + 5y \) at these vertices:
- At \( A(3, 0) \): \( Z = 3(3) + 5(0) = 9 \)
- At \( B\left(\frac{3}{2}, \frac{1}{2}\right) \): \( Z = 3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = 7 \)
- At \( C(0, 2) \): \( Z = 3(0) + 5(2) = 10 \)
Since we want to **maximize** \( Z \) in an unbounded region, we look at the behaviour of \( Z \) as \( x \) and \( y \) increase.
As the feasible region is unbounded and extends infinitely outwards, both \( x \) and \( y \) can be made arbitrarily large. Thus, the objective function \( Z = 3x + 5y \) can also grow infinitely.
Therefore, this linear programming problem has no finite maximum value; \( Z \) has an unbounded maximum.
In simple words: Since the shaded region is open-ended and goes on forever to the right and up, there is no limit to how big \( x \) and \( y \) can be. This means we can make \( Z \) as big as we want, so there is no maximum value.

Exam Tip: When maximizing in an unbounded region with \( \ge \) constraints, always explain clearly that the solution is unbounded and therefore no maximum exists.

Question 3. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains, 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs. 5 per kilogram and rice costs Rs. 4 per kilogram.
Answer:
Let \( x \) kg of bran and \( y \) kg of rice be mixed. Since the weights must be non-negative, \( x \ge 0, y \ge 0 \).
The objective function to minimize is the cost:
Minimize \( Z = 5x + 4y \)
Subject to the requirements:
1. Protein constraint (at least 88 grams):
\( 80x + 100y \ge 88 \implies 20x + 25y \ge 22 \)
2. Iron constraint (at least 36 milligrams):
\( 40x + 30y \ge 36 \implies 20x + 15y \ge 18 \)
Let us find the corner points of the unbounded feasible region:
- The line \( 20x + 25y = 22 \) has intercepts \( (1.1, 0) \) and \( (0, 0.88) \).
- The line \( 20x + 15y = 18 \) has intercepts \( (0.9, 0) \) and \( (0, 1.2) \).
Solving the two boundary equations simultaneously:
Subtracting the second equation from the first:
\( 10y = 4 \implies y = 0.4 \)
Substituting \( y = 0.4 \) into \( 20x + 15y = 18 \) gives:
\( 20x + 6 = 18 \implies 20x = 12 \implies x = 0.6 \).
So, the intersection point is \( (0.6, 0.4) \).
The corner points of the feasible region are:
- \( A(1.1, 0) \)
- \( B(0.6, 0.4) \)
- \( C(0, 1.2) \)
Evaluating \( Z = 5x + 4y \) at these points:
- At \( A(1.1, 0) \): \( Z = 5(1.1) + 4(0) = 5.5 \)
- At \( B(0.6, 0.4) \): \( Z = 5(0.6) + 4(0.4) = 4.6 \)
- At \( C(0, 1.2) \): \( Z = 5(0) + 4(1.2) = 4.8 \)
The minimum cost is Rs. 4.60, achieved by mixing 0.6 kg of bran and 0.4 kg of rice.
In simple words: To make the cereal at the lowest possible cost, mix 0.6 kg of bran and 0.4 kg of rice. This satisfies the daily nutrition targets for a minimum cost of Rs. 4.60.

Exam Tip: Be precise when simplifying the constraints to ensure you can find the coordinates of the corner points easily and accurately.

Question 4. A shopkeeper deals only in two items — tables and chairs. He has Rs. 6,000 to invest and a space to store at most 20 pieces. A table costs him Rs. 400 and a chair Rs. 250. He can sell a table at a profit of Rs. 25 and a chair at a profit of Rs. 40. Assume that he can sell all items that he buys. Using linear programming formulate the problem for maximum profit and solve it graphically.
Answer:
Let \( x \) and \( y \) be the number of tables and chairs purchased respectively. Since quantities cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to maximize is the total profit:
Maximize \( Z = 25x + 40y \)
Subject to the constraints:
1. Capital constraint (maximum Rs. 6,000):
\( 400x + 250y \le 6000 \implies 8x + 5y \le 120 \)
2. Storage constraint (maximum 20 pieces):
\( x + y \le 20 \)
Let us find the corner points of the bounded feasible region:
- The line \( 8x + 5y = 120 \) has intercepts \( (15, 0) \) and \( (0, 24) \).
- The line \( x + y = 20 \) has intercepts \( (20, 0) \) and \( (0, 20) \).
Solving the intersection of these two lines:
\( 8x + 5(20 - x) = 120 \implies 3x = 20 \implies x = \frac{20}{3} \approx 6.67 \)
Then \( y = \frac{40}{3} \approx 13.33 \).
The vertices of the feasible region are:
- \( O(0, 0) \)
- \( A(15, 0) \)
- \( B(6.67, 13.33) \)
- \( C(0, 20) \)
Let us evaluate \( Z = 25x + 40y \) at these vertices:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(15, 0) \): \( Z = 25(15) + 40(0) = 375 \)
- At \( B(6.67, 13.33) \): \( Z = 25(6.67) + 40(13.33) = 700 \)
- At \( C(0, 20) \): \( Z = 25(0) + 40(20) = 800 \)
Thus, the shopkeeper gets a maximum profit of Rs. 800 when he purchases only 20 chairs and 0 tables.
In simple words: Graphing the storage and money constraints shows us that the most profitable corner is buying only chairs. Purchasing 20 chairs costs Rs. 5000 (well within the Rs. 6000 limit) and generates the maximum profit of Rs. 800.

Exam Tip: Always evaluate the objective function at every corner point to mathematically justify the optimal choice, even if the result suggests zero of one item.

Page 72

Questions for Self Evaluation

Question 1. Solve the following linear programming problem graphically : Maximize z = x – 7y +190 subject to the constraints x + y  8, x  5, y  5, x + y  4, x  0, y  0 .
Answer:
Let us identify the vertices (corner points) of the bounded feasible region defined by the constraints:
- \( x + y \le 8 \)
- \( x \le 5 \)
- \( y \le 5 \)
- \( x + y \ge 4 \)
- \( x \ge 0, \quad y \ge 0 \)
The boundary lines are:
- \( x + y = 8 \)
- \( x = 5 \)
- \( y = 5 \)
- \( x + y = 4 \)
The corner points are:
- \( A(4, 0) \)
- \( B(5, 0) \)
- \( C(5, 3) \)
- \( D(3, 5) \)
- \( E(0, 5) \)
- \( F(0, 4) \)
Evaluating the objective function \( Z = x - 7y + 190 \) at each corner point:
- At \( A(4, 0) \): \( Z = 4 - 7(0) + 190 = 194 \)
- At \( B(5, 0) \): \( Z = 5 - 7(0) + 190 = 195 \)
- At \( C(5, 3) \): \( Z = 5 - 7(3) + 190 = 174 \)
- At \( D(3, 5) \): \( Z = 3 - 7(5) + 190 = 158 \)
- At \( E(0, 5) \): \( Z = 0 - 7(5) + 190 = 155 \)
- At \( F(0, 4) \): \( Z = 0 - 7(4) + 190 = 162 \)
Comparing these values, the maximum value of \( Z \) is \( 195 \) at the point \( (5, 0) \).
In simple words: Plot all constraint lines on a graph. Find the common shaded area, then check the value of \( Z \) at each corner. The highest value is 195, which is located at the corner point \( (5, 0) \).

Exam Tip: Be careful to read the signs in the inequalities. Draw a quick arrow on each boundary line indicating which side of the line is favored (above or below) to identify the correct feasible region.

Question 2. Solve the following linear programming problem graphically : Maximize z = 3x + 5y subject to the constraints x + y  2, x + 3y  3, x  0, y  0 .
Answer:
The boundary lines are:
1. \( x + y = 2 \)
2. \( x + 3y = 3 \)
Since both constraints are of the type \( \ge \), the feasible region is unbounded and extends infinitely in the first quadrant away from the origin.
The corner points of this unbounded region are:
- \( A(3, 0) \)
- \( B\left(\frac{3}{2}, \frac{1}{2}\right) \)
- \( C(0, 2) \)
Let us evaluate \( Z = 3x + 5y \) at these vertices:
- At \( A(3, 0) \): \( Z = 3(3) + 5(0) = 9 \)
- At \( B\left(\frac{3}{2}, \frac{1}{2}\right) \): \( Z = 3\left(\frac{3}{2}\right) + 5\left(\frac{1}{2}\right) = 7 \)
- At \( C(0, 2) \): \( Z = 3(0) + 5(2) = 10 \)
Since we want to **maximize** \( Z \) in an unbounded region, we look at the behaviour of \( Z \) as \( x \) and \( y \) increase.
As the feasible region is unbounded and extends infinitely outwards, both \( x \) and \( y \) can be made arbitrarily large. Thus, the objective function \( Z = 3x + 5y \) can also grow infinitely.
Therefore, this linear programming problem has no finite maximum value; \( Z \) has an unbounded maximum.
In simple words: Since the shaded region is open-ended and goes on forever to the right and up, there is no limit to how big \( x \) and \( y \) can be. This means we can make \( Z \) as big as we want, so there is no maximum value.

Exam Tip: When maximizing in an unbounded region with \( \ge \) constraints, always explain clearly that the solution is unbounded and therefore no maximum exists.

Question 3. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains, 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs. 5 per kilogram and rice costs Rs. 4 per kilogram.
Answer:
Let \( x \) kg of bran and \( y \) kg of rice be mixed. Since the weights must be non-negative, \( x \ge 0, y \ge 0 \).
The objective function to minimize is the cost:
Minimize \( Z = 5x + 4y \)
Subject to the requirements:
1. Protein constraint (at least 88 grams):
\( 80x + 100y \ge 88 \implies 20x + 25y \ge 22 \)
2. Iron constraint (at least 36 milligrams):
\( 40x + 30y \ge 36 \implies 20x + 15y \ge 18 \)
Let us find the corner points of the unbounded feasible region:
- The line \( 20x + 25y = 22 \) has intercepts \( (1.1, 0) \) and \( (0, 0.88) \).
- The line \( 20x + 15y = 18 \) has intercepts \( (0.9, 0) \) and \( (0, 1.2) \).
Solving the two boundary equations simultaneously:
Subtracting the second equation from the first:
\( 10y = 4 \implies y = 0.4 \)
Substituting \( y = 0.4 \) into \( 20x + 15y = 18 \) gives:
\( 20x + 6 = 18 \implies 20x = 12 \implies x = 0.6 \).
So, the intersection point is \( (0.6, 0.4) \).
The corner points of the feasible region are:
- \( A(1.1, 0) \)
- \( B(0.6, 0.4) \)
- \( C(0, 1.2) \)
Evaluating \( Z = 5x + 4y \) at these points:
- At \( A(1.1, 0) \): \( Z = 5(1.1) + 4(0) = 5.5 \)
- At \( B(0.6, 0.4) \): \( Z = 5(0.6) + 4(0.4) = 4.6 \)
- At \( C(0, 1.2) \): \( Z = 5(0) + 4(1.2) = 4.8 \)
The minimum cost is Rs. 4.60, achieved by mixing 0.6 kg of bran and 0.4 kg of rice.
In simple words: To make the cereal at the lowest possible cost, mix 0.6 kg of bran and 0.4 kg of rice. This satisfies the daily nutrition targets for a minimum cost of Rs. 4.60.

Exam Tip: Be precise when simplifying the constraints to ensure you can find the coordinates of the corner points easily and accurately.

Question 4. A shopkeeper deals only in two items — tables and chairs. He has Rs. 6,000 to invest and a space to store at most 20 pieces. A table costs him Rs. 400 and a chair Rs. 250. He can sell a table at a profit of Rs. 25 and a chair at a profit of Rs. 40. Assume that he can sell all items that he buys. Using linear programming formulate the problem for maximum profit and solve it graphically.
Answer:
Let \( x \) and \( y \) be the number of tables and chairs purchased respectively. Since quantities cannot be negative:
\( x \ge 0, \quad y \ge 0 \)
The objective function to maximize is the total profit:
Maximize \( Z = 25x + 40y \)
Subject to the constraints:
1. Capital constraint (maximum Rs. 6,000):
\( 400x + 250y \le 6000 \implies 8x + 5y \le 120 \)
2. Storage constraint (maximum 20 pieces):
\( x + y \le 20 \)
Let us find the corner points of the bounded feasible region:
- The line \( 8x + 5y = 120 \) has intercepts \( (15, 0) \) and \( (0, 24) \).
- The line \( x + y = 20 \) has intercepts \( (20, 0) \) and \( (0, 20) \).
Solving the intersection of these two lines:
\( 8x + 5(20 - x) = 120 \implies 3x = 20 \implies x = \frac{20}{3} \approx 6.67 \)
Then \( y = \frac{40}{3} \approx 13.33 \).
The vertices of the feasible region are:
- \( O(0, 0) \)
- \( A(15, 0) \)
- \( B(6.67, 13.33) \)
- \( C(0, 20) \)
Let us evaluate \( Z = 25x + 40y \) at these vertices:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(15, 0) \): \( Z = 25(15) + 40(0) = 375 \)
- At \( B(6.67, 13.33) \): \( Z = 25(6.67) + 40(13.33) = 700 \)
- At \( C(0, 20) \): \( Z = 25(0) + 40(20) = 800 \)
Thus, the shopkeeper gets a maximum profit of Rs. 800 when he purchases only 20 chairs and 0 tables.
In simple words: Graphing the storage and money constraints shows us that the most profitable corner is buying only chairs. Purchasing 20 chairs costs Rs. 5000 (well within the Rs. 6000 limit) and generates the maximum profit of Rs. 800.

Exam Tip: Always evaluate the objective function at every corner point to mathematically justify the optimal choice, even if the result suggests zero of one item.

Question 5. A small firm manufactures items A and B. The total number of items A and B it can manufacture a day is at most 24. Item A takes one hour to make while item B takes only half an hour. The maximum time available per day is 16 hours. If the profit on one unit of item A be Rs. 300 and one unit of item B be Rs. 160, how many of each type of item be produced to maximize the profit ? Solve the problem graphically.
Answer:
Let \( x \) and \( y \) be the daily production numbers for items A and B respectively. Since counts cannot be negative, we have \( x \ge 0, y \ge 0 \).
The objective function to maximize is the total profit:
Maximize \( Z = 300x + 160y \)
Subject to the constraints:
1. Daily manufacturing limit (at most 24 items):
\( x + y \le 24 \)
2. Time availability limit (at most 16 hours):
\( 1x + 0.5y \le 16 \implies 2x + y \le 32 \)
Let us find the boundary lines to locate the vertices of the bounded feasible region:
- The line \( x + y = 24 \) passes through \( (24, 0) \) and \( (0, 24) \).
- The line \( 2x + y = 32 \) passes through \( (16, 0) \) and \( (0, 32) \).
Solving these equations simultaneously to find the intersection point:
Subtracting the first equation from the second:
\( x = 8 \)
Substituting \( x = 8 \) into \( x + y = 24 \) gives \( y = 16 \). So, the intersection is \( B(8, 16) \).
The vertices of the bounded feasible region are:
- \( O(0, 0) \)
- \( A(16, 0) \)
- \( B(8, 16) \)
- \( C(0, 24) \)
Let us evaluate \( Z = 300x + 160y \) at each of these corner points:
- At \( O(0, 0) \): \( Z = 0 \)
- At \( A(16, 0) \): \( Z = 300(16) + 160(0) = 4800 \)
- At \( B(8, 16) \): \( Z = 300(8) + 160(16) = 2400 + 2560 = 4960 \)
- At \( C(0, 24) \): \( Z = 300(0) + 160(24) = 3840 \)
Comparing these, the maximum profit is Rs. 4,960 when 8 units of item A and 16 units of item B are produced daily.
In simple words: Draw both constraint lines on a graph. Test the corner coordinates of the shared shaded area in the profit function. Producing 8 units of item A and 16 units of item B gives the highest possible profit of Rs. 4,960.

Exam Tip: Be sure to write out the simplified form of the equations to make plotting and intersection solving straightforward and neat.

Question 6. A chemist requires 10, 12 and 12 units of chemicals A, B and C respectively for his analysis. A liquid product contains 5, 2, and 1 units of A, B and C respectively and it costs Rs. 3 per jar. A dry product contains 1, 2, and 4 units of A, B and C per carton and costs Rs. 2 per carton. How many of each should he purchase in order to minimize the cost and meet the requirement ?
Answer:
Let the chemist purchase \( x \) jars of the liquid product and \( y \) cartons of the dry product. Since quantities cannot be negative, we have \( x \ge 0, y \ge 0 \).
The objective function to minimize is the cost:
Minimize \( Z = 3x + 2y \)
Subject to the chemical requirement constraints:
1. Chemical A requirement (at least 10 units):
\( 5x + y \ge 10 \)
2. Chemical B requirement (at least 12 units):
\( 2x + 2y \ge 12 \implies x + y \ge 6 \)
3. Chemical C requirement (at least 12 units):
\( x + 4y \ge 12 \)
Let us plot the boundary lines to determine the vertices of the unbounded feasible region:
- The line \( 5x + y = 10 \) passes through \( (2, 0) \) and \( (0, 10) \).
- The line \( x + y = 6 \) passes through \( (6, 0) \) and \( (0, 6) \).
- The line \( x + 4y = 12 \) passes through \( (12, 0) \) and \( (0, 3) \).
Solving the intersections of adjacent lines:
- Intersection of \( 5x + y = 10 \) and \( x + y = 6 \) is \( C(1, 5) \).
- Intersection of \( x + y = 6 \) and \( x + 4y = 12 \) is \( B(4, 2) \).
The vertices of the unbounded feasible region are:
- \( A(12, 0) \)
- \( B(4, 2) \)
- \( C(1, 5) \)
- \( D(0, 10) \)
Let us evaluate the cost \( Z = 3x + 2y \) at each vertex:
- At \( A(12, 0) \): \( Z = 3(12) + 2(0) = 36 \)
- At \( B(4, 2) \): \( Z = 3(4) + 2(2) = 16 \)
- At \( C(1, 5) \): \( Z = 3(1) + 2(5) = 13 \)
- At \( D(0, 10) \): \( Z = 3(0) + 2(10) = 20 \)
Since the region is unbounded, we verify that the open half-plane \( 3x + 2y < 13 \) does not have any points in common with the feasible region.
Thus, the minimum cost is Rs. 13, which is achieved by purchasing 1 jar of the liquid product and 5 cartons of the dry product.
In simple words: Setup the three inequalities on a graph and locate the open-ended shaded region. Comparing the cost at the corner points shows that buying 1 jar of liquid product and 5 cartons of dry product is the cheapest option at Rs. 13.

Exam Tip: In unbounded minimization problems, always double-check the corner points to ensure that the candidate minimum vertex actually satisfies all constraint equations.

Question 7. A person wants to invest at most Rs. 18,000 in Bonds A and B. According to the rules, he has to invest at least Rs. 4,000 in Bond A and at least Rs. 5,000 in Bond B. If the rate of interest on bond A is 9 % per annum and the rate of interest on bond B is 11 % per annum , how much money should he invest to earn maximum yearly income ?
Answer:
Let Rs. \( x \) and Rs. \( y \) be invested in Bond A and Bond B respectively. Since investments cannot be negative, we have \( x \ge 0, y \ge 0 \).
The objective function to maximize is the annual interest income:
Maximize \( Z = 0.09x + 0.11y \)
Subject to the constraints:
1. Maximum total investment: \( x + y \le 18000 \)
2. Minimum investment in Bond A: \( x \ge 4000 \)
3. Minimum investment in Bond B: \( y \ge 5000 \)
The boundary lines are:
- \( x + y = 18000 \)
- \( x = 4000 \)
- \( y = 5000 \)
The corner points of the bounded triangular feasible region are:
- \( P(4000, 5000) \)
- \( Q(13000, 5000) \) [intersection of \( y = 5000 \) and \( x + y = 18000 \)]
- \( R(4000, 14000) \) [intersection of \( x = 4000 \) and \( x + y = 18000 \)]
Let us evaluate \( Z = 0.09x + 0.11y \) at each corner point:
- At \( P(4000, 5000) \): \( Z = 0.09(4000) + 0.11(5000) = 360 + 550 = 910 \)
- At \( Q(13000, 5000) \): \( Z = 0.09(13000) + 0.11(5000) = 1170 + 550 = 1720 \)
- At \( R(4000, 14000) \): \( Z = 0.09(4000) + 0.11(14000) = 360 + 1540 = 1900 \)
Comparing these interest values, the maximum yearly income is Rs. 1,900, which occurs when Rs. 4,000 is invested in Bond A and Rs. 14,000 is invested in Bond B.
In simple words: Since Bond B pays a higher interest rate (11% compared to 9%), the investor should put as much money as possible into Bond B after satisfying the minimum Rs. 4,000 requirement for Bond A. This results in putting Rs. 4,000 into A and Rs. 14,000 into B.

Exam Tip: For simple allocation problems where one option clearly has a higher rate of return, the optimal solution will allocate the maximum possible amount allowed by the constraints to the higher-yield option.

Question 8. Two tailors A and B earn Rs. 150 and Rs. 200 per day respectively. A can stitch 6 shirts and 4 pants while B can stitch 10 shirts and 4 pants per day. How many days shall each work if it is desired to to stitch at least 60 shirts and 32 pants at a minimum labourcost.
Answer:
Let tailor A work for \( x \) days and tailor B work for \( y \) days. Since the number of days cannot be negative, we have \( x \ge 0, y \ge 0 \).
The objective function to minimize is the total labor cost:
Minimize \( Z = 150x + 200y \)
Subject to the constraints:
1. Shirts requirement (at least 60 shirts):
\( 6x + 10y \ge 60 \implies 3x + 5y \ge 30 \)
2. Pants requirement (at least 32 pants):
\( 4x + 4y \ge 32 \implies x + y \ge 8 \)
Let us plot the boundary lines to find the corner points of the unbounded feasible region:
- The line \( 3x + 5y = 30 \) has intercepts \( (10, 0) \) and \( (0, 6) \).
- The line \( x + y = 8 \) has intercepts \( (8, 0) \) and \( (0, 8) \).
Solving the two boundary equations simultaneously to find their intersection point:
Multiply \( x + y = 8 \) by 3:
\( 3x + 3y = 24 \)
Subtracting this from \( 3x + 5y = 30 \) gives:
\( 2y = 6 \implies y = 3 \)
Substituting \( y = 3 \) back into \( x + y = 8 \) gives \( x = 5 \). So, the intersection point is \( B(5, 3) \).
The corner points of the feasible region are:
- \( A(10, 0) \)
- \( B(5, 3) \)
- \( C(0, 8) \)
Let us evaluate \( Z = 150x + 200y \) at these vertices:
- At \( A(10, 0) \): \( Z = 150(10) + 200(0) = 1500 \)
- At \( B(5, 3) \): \( Z = 150(5) + 200(3) = 1350 \)
- At \( C(0, 8) \): \( Z = 150(0) + 200(8) = 1600 \)
Since the feasible region is unbounded, we verify that the open half-plane \( 150x + 200y < 1350 \) does not overlap with the feasible region.
Therefore, the minimum labor cost is Rs. 1,350, which occurs when tailor A works for 5 days and tailor B works for 3 days.
In simple words: We setup the equations on a coordinate graph to find the open shaded region. Testing the corners reveals that having tailor A work for 5 days and tailor B work for 3 days results in the lowest labor cost of Rs. 1,350.

Exam Tip: Ensure that you simplify the constraint inequalities (like dividing by common factors) before plotting them, as this keeps the calculations clear and easy to manage.