CBSE Class 12 Mathematics Linear Programming Worksheet Set 02

MULTIPLE CHOICE QUESTIONS

Question. The corner points of the feasible region determined by the system of linear constraints are (0, 0), (0,40), (20,40),(60,20),(60,0).The objective function is Compare the quantity in Column A and Column B
Column A              Column B
Maximum of Z           325
(a) The quantity in column A is greater
(b) The quantity in column B is greater
(c) The two quantities are equal.
(d) The relationship cannot be determined on the basis of the information supplied.
Answer : B

Question. The feasible solution for a LPP is shown in given figure. Let Z=3x-4y be the objective function. Minimum of Z occurs at

(a) (0,0)
(b) (0,8)
(c) (5,0)
(d) (4,10)
Answer : B

Question. Corner points of the feasible region determined by the system of linear constraints are (0,3),(1,1) and (3,0). Let Z= px+qy, where p, q>0. Condition on p and q so that the minimum of Z occurs at (3,0) and (1,1) is
(a) p=2q
(b) p=q/2
(c) p=3q
(d) p=q
Answer : B

Question. The set of all feasible solutions of a LPP is a ____ set.
(a) Concave
(b) Convex
(c) Feasible
(d) None of these
Answer : A

Question. Corner points of the feasible region for an LPP are (0,2), (3,0), (6,0), (6,8) and (0,5). Let F=4x+6y be the objective function. Maximum of F – Minimum of F =
(a) 60
(b) 48
(c) 42
(d) 18
Answer : A

Question. In a LPP, if the objective function Z = ax+by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same……….value.
(a) minimum
(b) maximum
(c) zero
(d) none of these
Answer : B

Question. In the feasible region for a LPP is ………, then the optimal value of the objective function Z = ax+bymayormaynot exist.
(a) bounded
(b) unbounded
(c) in circled form
(d) in squared form
Answer : B

Question. A linear programming problem is one that is concerned with finding the …A … of a linear function called …B… function of several values (say x and y), subject to the conditions that the variables are …C… and satisfy set of linear inequalities called linear constraints.
(a) Objective, optimal value, negative
(b) Optimal value, objective, negative
(c) Optimal value, objective, nonnegative
(d) Objective, optimal value, nonnegative
Answer : C

Question. Maximum value of the objective function Z = ax+by in a LPP always occurs at only one corner point of the feasible region.
(a) true
(b) false
(c) can’t say
(d) partially true
Answer : B

Question. Region represented by x≥0,y≥0 is:
(a) First quadrant
(b) Second quadrant
(c) Third quadrant
(d) Fourth quadrant
Answer : A

Question. Z =3x + 4y, Subject to the constraints x+y 1, x,y ≥0. the shaded region shown in the figure as OAB is bounded and thecoordinatesof corner points O, A and B are (0,0),(1,0) and (0,1), respectively.

The maximum value of Z is 2.
(a) true
(b) false
(c) can’t say
(d) partially true
Answer : B

Question. The feasible region for an LPP is shown shaded in the figure. Let Z = 3x-4y be objective function. Maximum value of Z is:

(a) 0
(b) 8
(c) 12
(d) -18
Answer : A

Question. The maximum value of Z = 4x+3y, if the feasible region for an LPP is as shown below, is

(a) 112
(b) 100
(c) 72
(d) 110
Answer : A

Question. The feasible region for an LPP is shown shaded in the figure. Let Z = 4x-3y be objective function. Maximum value of Z is:

(a) 0
(b) 8
(c) 30
(d) -18
Answer : C

Question. In the given figure, the feasible region for a LPP is shown. Find the maximum and minimum value of Z = x + 2y.

(a) 8, 3.2
(b) 9, 3.14
(c) 9, 4
(d) none of these
Answer : B

Question. The linear programming problem minimize Z= 3x+2y,subject to constraints x+y8, 3x+5y 15, x,y ≥0, has
(a) One solution
(b) No feasible solution
(c) Two solutions
(d) Infinitely many solutions
Answer : B

Question. The graph of the inequality 2x+3y > 6 is:
(a) half plane that contains the origin
(b) half plane that neither contains the origin nor the points of the line 2x+3y =6
(c) whole XOY-plane excluding the points on the line 2x+3y =6
(d) entire XOY-plane
Answer : B

Question. Of all the points of the feasible region for maximum or minimum of objective function the points
(a) Inside the feasible region
(b) At the boundary line of the feasible region
(c) Vertex point of the boundary of the feasible region
(d) None of these
Answer : C

Question. The maximum value of the object function Z = 5x + 10 y subject to the constraints x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x ≥ 0, y ≥ 0 is
(a) 300
(b) 600
(c) 400
(d) 800
Answer : B

Question. Z = 6x + 21 y, subject to x + 2y ≥ 3, x + 4y ≥ 4, 3x + y ≥ 3, x ≥ 0, y ≥ 0. The minimum value of Z occurs at
(a) (4, 0)
(b) (28, 8)
(c) (2,2/7)
(d) (0, 3)
Answer : C

Question. Shape of the feasible region formed by the following constraints x + y ≤ 2, x + y ≥ 5, x ≥ 0, y ≥ 0
(a) No feasible region
(b) Triangular region
(c) Unbounded solution
(d) Trapezium
Answer : A

Question. Maximize Z = 4x + 6y, subject to 3x + 2y ≤ 12, x + y ≥ 4, x, y ≥ 0.
(a) 16 at (4, 0)
(b) 24 at (0, 4)
(c) 24 at (6, 0)
(d) 36 at (0, 6)
Answer : D

Question. Feasible region for an LPP shown shaded in the following figure. Minimum of Z = 4x+3y occurs at the point:

(a) (0,8)
(b) (2,5)
(c) (4,3)
(d) (9,0)
Answer : B

Question. The region represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 0, x ≥ 0, y ≥ 0 is
(a) unbounded
(b) a polygon
(c) exterior of a triangle
(d) None of these
Answer : D

Question. Minimize Z = 13x – 15y subject to the constraints : x + y ≤ 7, 2x – 3y + 6 ≥ 0 , x ≥ 0, y ≥ 0.
(a) -23
(b) -32
(c) -30
(d) -34
Answer : C

LINEAR PROGRAMMING (LPP)

Question 1. A small manufactured has employed 5 skilled me n & 10 semi-skilled men and makes an article in two qualities deluxe model and an ordinary model. The making of a deluxe model requires 2 ℎ𝑟𝑠 work by skilled man and 2 ℎ𝑟 work by a semi-skilled man. The ordinary model requires 1 ℎ𝑟 by a skilled man and 3ℎ𝑟𝑠 by a semi-skilled man. By union rules no man may work more than 8 ℎ𝑟𝑠 per day. The profit on a deluxe model is 𝑅𝑠 15 and an ordinary model is 𝑅𝑠 10. How many of each type should be made to maximize his daily profit ?
Answer:
Let \( x \) and \( y \) denote the quantities of deluxe models and ordinary models produced daily, respectively.
The skilled labor constraint is determined by the 5 skilled workers. Since each worker can work at most 8 hours per day, the total skilled labor hours available daily is \( 5 \times 8 = 40 \) hours. Thus, we have:
\[ 2x + y \le 40 \quad \dots (1) \]
Similarly, for the 10 semi-skilled workers, each working at most 8 hours daily, the total semi-skilled labor hours available is \( 10 \times 8 = 80 \) hours. This gives:
\[ 2x + 3y \le 80 \quad \dots (2) \]
Additionally, the non-negativity constraints are:
\[ x \ge 0, \quad y \ge 0 \]
Our objective is to maximize the daily profit function:
\[ Z = 15x + 10y \]
To find the optimal solution, we plot the boundary lines of these inequalities on a graph and identify the feasible region. This region is a polygon \( OABC \) with the corner points \( O(0,0) \), \( C(20,0) \), \( B(10,20) \), and \( A(0, 80/3) \).

We now evaluate the objective function \( Z \) at each corner point of the feasible region:
- At \( O(0,0) \): \( Z = 15(0) + 10(0) = 0 \)
- At \( C(20,0) \): \( Z = 15(20) + 10(0) = 300 \)
- At \( B(10,20) \): \( Z = 15(10) + 10(20) = 350 \)
- At \( A(0, 80/3) \): \( Z = 15(0) + 10(80/3) = \frac{800}{3} \approx 266.67 \)

The maximum profit is Rs. 350, which occurs at the corner point \( B(10, 20) \).
Therefore, the manufacturer should produce 10 deluxe models and 20 ordinary models daily to achieve the maximum profit of Rs. 350.
In simple words: To make the most profit under the given work-hour limits, the factory should produce 10 deluxe models and 20 ordinary models every day. This will bring in a maximum daily profit of Rs. 350.

Exam Tip: Be careful when calculating the total daily hours available for each class of labor. Multiply the number of workers by the maximum working hours per day to set up your inequality constraints correctly.

Question 2. An aeroplane can carry a maximum of 200 passengers. A profit of Rs.1000 is made on each executive class tickets and a profit of Rs. 600 is made on each economy class tickets. The airline reserves at least 20 seats for executive class. However, at least4 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 what is the max profit ?
Answer:
Let \( x \) represent the number of executive class tickets sold and \( y \) represent the number of economy class tickets sold.
We formulate the linear programming problem as follows:
1) Since the seating capacity of the aircraft is limited to 200, we have:
\[ x + y \le 200 \]
2) At least 20 seats must be reserved for the executive class:
\[ x \ge 20 \]
3) The number of economy class tickets sold is at least 4 times the number of executive class tickets:
\[ y \ge 4x \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
The total profit to be maximized is given by the objective function:
\[ Z = 1000x + 600y \]
Plotting these inequalities on the graph below, we determine the bounded feasible region as the triangle \( ABC \):

Next, we calculate the values of the objective function \( Z \) at each corner point of the feasible region:
- At \( A(20, 180) \): \( Z = 1000(20) + 600(180) = 20000 + 108000 = 128000 \)
- At \( B(40, 160) \): \( Z = 1000(40) + 600(160) = 40000 + 96000 = 136000 \)
- At \( C(20, 80) \): \( Z = 1000(20) + 600(80) = 20000 + 48000 = 68000 \)

The maximum value of the objective function is Rs. 1,36,000, which occurs at the corner point \( B(40, 160) \).
Therefore, the airline should sell 40 executive class tickets and 160 economy class tickets to obtain the maximum profit of Rs. 1,36,000.
In simple words: To get the highest possible profit of Rs. 1,36,000, the airline needs to sell exactly 40 executive class tickets and 160 economy class tickets. This plan perfectly fits the plane's capacity and seat reservations.

Exam Tip: Always double check the vertices of the feasible region by solving the pairs of boundary lines algebraically. A small graphing error can lead to incorrect corner points and an incorrect maximum value.

Question 3. A toy company manufactures two types of dolls A and B. market tests and available resources indicated that the combined production level should not exceed 1200 dolls per week and the demand for doll of type B is at most half of that for dolls of type A. Further the production level of dolls of type A can exceed three times the production of dolls of other type by at most 600 units. If the company makes profit of Rs.12 and Rs.16 per doll respectively on doll A and B. how many of each should be produced to max the profit ?
Answer:
Let \( x \) and \( y \) represent the number of type A and type B dolls produced per week, respectively.
We can formulate the linear programming problem as follows:
Objective function to maximize the weekly profit is:
\[ Z = 12x + 16y \]
Subject to the following constraints:
1) The combined production level does not exceed 1200 dolls per week:
\[ x + y \le 1200 \]
2) The demand for type B dolls is at most half of that for type A dolls:
\[ y \le \frac{x}{2} \implies x \ge 2y \]
3) The production of type A dolls exceeds three times the production of type B dolls by at most 600 units:
\[ x - 3y \le 600 \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we find the shaded feasible region as shown below:

We now evaluate the objective function \( Z = 12x + 16y \) at each corner point of the feasible region:

The maximum profit is Rs. 16,000, which is achieved when producing 800 dolls of type A and 400 dolls of type B.
Therefore, the company should produce 800 type A dolls and 400 type B dolls per week to obtain the maximum profit of Rs. 16,000.
In simple words: To maximize weekly profits at Rs. 16,000, the company should manufacture exactly 800 dolls of type A and 400 dolls of type B. This matches the resources and customer demand patterns perfectly.

Exam Tip: Be careful when simplifying inequalities. For example, rewriting \( y \le \frac{x}{2} \) as \( x \ge 2y \) helps you accurately plot the boundary line and correctly determine which side of the line is the feasible region.

Question 4. A factory owner purchases two types of machines A and B for his factory. The requirements and limitation for the machines are as follows :

He has an area of 7600 sq. m available and 72 skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output ?
Answer:
Let \( x \) and \( y \) represent the number of Machine A and Machine B purchased, respectively.
The LPP is formulated to maximize the daily output function:
\[ Z = 60x + 40y \]
Subject to the following constraints (based on the total available area of 9000 sq. m as shown in the mathematical solution):
1) Area constraint:
\[ 1000x + 1200y \le 9000 \implies 5x + 6y \le 45 \quad \dots (1) \]
2) Skilled labor constraint:
\[ 12x + 8y \le 72 \implies 3x + 2y \le 18 \quad \dots (2) \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
The feasible region \( OABC \) bounded by these constraints is plotted below:

We now evaluate the objective function \( Z = 60x + 40y \) at each corner point of the feasible region:

Although the mathematical maximum of 360 units is achieved at the intersection point \( B(2.25, 5.625) \), the number of machines purchased must be positive integers. Looking at the integer corner point \( C(6,0) \), the maximum output is also 360 units.
Therefore, the factory owner should buy 6 machines of type A to maximize the daily output.
In simple words: To get the highest possible output without using fractional machines, the owner should buy exactly 6 machines of Type A (and zero of Type B). This keeps the total output at a maximum of 360 units daily.

Exam Tip: In practical/real-world problems, fractional values of variables (like buying 2.25 machines) are not possible. Always look for nearby integer coordinates within the feasible region to find the true practical optimal solution.

Question 5. A library has to accommodate two different types of books on a shelf. The books are 6 cm and 4 cm thick and weight 1 kg and 1½ kg each respectively. The shelf is 96 cm long and at most can support a weight of 21 kg. How should be the shelf be find with books if two types in order to include the greater number of books ?
Answer:
Let \( x \) represent the number of 1st type books and \( y \) represent the number of 2nd type books.
We formulate the linear programming problem to maximize the total number of books:
\[ Z = x + y \]
Subject to the following constraints:
1) Space/thickness constraint (since the shelf is 96 cm long):
\[ 6x + 4y \le 96 \implies 3x + 2y \le 48 \quad \dots (1) \]
2) Weight constraint (since the shelf can support at most 21 kg):
\[ x + 1.5y \le 21 \implies 2x + 3y \le 42 \quad \dots (2) \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
We graph these inequalities to identify the feasible region, which is bounded by the corner points \( O(0,0) \), \( (16,0) \), \( (12,6) \), and \( (0,14) \):

We evaluate the total number of books \( Z = x + y \) at the corner points:
- At \( (16,0) \): \( Z = 16 + 0 = 16 \)
- At \( (0,14) \): \( Z = 0 + 14 = 14 \)
- At \( (12,6) \): \( Z = 12 + 6 = 18 \)

The maximum value of \( Z \) is 18 books, which is achieved at the intersection point \( (12,6) \).
Thus, the library should place 12 books of the first type and 6 books of the second type on the shelf to maximize the number of books.
In simple words: To fit the most books on the shelf without exceeding the length or weight limit, the librarian should put 12 of the first type of books and 6 of the second type of books. This allows for a total of 18 books on the shelf.

Exam Tip: When formulating constraints with fractional coefficients like \( 1\frac{1}{2} \) kg, convert them to improper fractions \( \frac{3}{2} \) or decimals like \( 1.5 \) and clear the fractions to avoid computational errors in subsequent steps.

Question 6. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 𝑔𝑚 of protein and at least 36 milligram of iron knowing that bran contains 80 grams of protein and 40 milligram of iron per kilogram and that rice contains 100 grams of protein and 30 milligram of iron per kilogram. Find the minimum cost Rs.5 per kg and rice cost 𝑅𝑠. 4 𝑝𝑒 r 𝑘𝑔.
Answer:
Let \( x \) represent the amount (in kg) of bran and \( y \) represent the amount (in kg) of rice in the cereal mixture.
We formulate the linear programming problem to minimize the cost of the mixture:
\[ \text{Minimize } Z = 5x + 4y \]
Subject to the following nutritional constraints:
1) Protein constraint (at least 88 g, given bran contains 80 g/kg and rice contains 100 g/kg):
\[ 80x + 100y \ge 88 \implies 20x + 25y \ge 22 \quad \dots (1) \]
2) Iron constraint (at least 36 mg, given bran contains 40 mg/kg and rice contains 30 mg/kg):
\[ 40x + 30y \ge 36 \implies 20x + 15y \ge 18 \quad \dots (2) \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we find that the feasible region is unbounded and has the corner points \( A(1.1, 0) \), \( E(0.6, 0.4) \), and \( D(0, 1.2) \):

We evaluate the total cost \( Z = 5x + 4y \) at each corner point of this unbounded feasible region:

The minimum cost is Rs. 4.60 per kg, which occurs when using 0.6 kg of bran and 0.4 kg of rice.
Therefore, the company should mix 0.6 kg of bran and 0.4 kg of rice to minimize the cost of the cereal mixture to Rs. 4.60 per kg.
In simple words: To make the cereal mixture as cheaply as possible while still meeting the minimum nutrition guidelines, the factory should mix 0.6 kg of bran and 0.4 kg of rice. This will cost exactly Rs. 4.60 per kg.

Exam Tip: For an unbounded feasible region in minimization LPP, always evaluate the objective function at all corner points. Since there is no upper bound, the minimum value will lie on one of these vertices.

Question 7. A fruit grower can use two types of fertilizer in his garden brand P and brand Q. The amount (in kg) of nitrogen, phosphoric acid, potash and chlorine in a bag of watch brand are given in the table :-

Tests indicate the garden needs at least 240kg of phosphoric acid, at least 270 kg of potash and at most 310 kg of chlorine. If the grower wants to minimize the amount of nitrogen added to the garden. How many bags of each brand should be used? What is the minimum amount of nitrogen added ?
Answer:
Let \( x \) denote the number of bags of Brand P fertilizer used, and let \( y \) denote the number of bags of Brand Q fertilizer used.
We formulate the linear programming problem to minimize the total nitrogen added:
\[ \text{Minimize } Z = 3x + 3.5y \]
Subject to the following constraints:
1) Phosphoric acid constraint (at least 240 kg):
\[ x + 2y \ge 240 \]
2) Potash constraint (at least 270 kg):
\[ 3x + 1.5y \ge 270 \]
3) Chlorine constraint (at most 310 kg):
\[ 1.5x + 2y \le 310 \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we locate the feasible region bounded by the corner points \( P(20, 140) \), \( Q(40, 100) \), and \( R(140, 50) \):

We evaluate the total nitrogen function \( Z = 3x + 3.5y \) at each corner point:

The minimum nitrogen content is 470 kg, which is achieved at the corner point \( Q(40, 100) \).
Therefore, the fruit grower should use 40 bags of brand P and 100 bags of brand Q to minimize the nitrogen content to 470 kg.
In simple words: To keep the amount of nitrogen in the garden as low as possible while still getting enough potash and phosphoric acid, the grower should use exactly 40 bags of brand P and 100 bags of brand Q. This keeps the total nitrogen at a minimum of 470 kg.

Exam Tip: Pay special attention to the direction of inequalities. Terms like "at least" translate to \( \ge \), while terms like "at most" translate to \( \le \). Reversing these will completely change the feasible region.

Question 8. There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing soil conditions, a farmer finds that she needs at least 14 kg of nitrogen 14 kg of phosphoric acid for crop. If F1 costs Rs.6 / kg and F2 costs Rs.5/kg. Determine how much of each type of fertilizer should be used so that the cost is minimum ?
Answer:
Let \( x \) kg of fertilizer \( F_1 \) and \( y \) kg of fertilizer \( F_2 \) be purchased. We construct the following table from the given conditions:

Our objective is to minimize the total cost of the fertilizers:
\[ \text{Minimize } Z = 6x + 5y \]
Subject to the following constraints:
1) Nitrogen requirement (at least 14 kg):
\[ 10\% \text{ of } x + 5\% \text{ of } y \ge 14 \implies \frac{10x}{100} + \frac{5y}{100} \ge 14 \implies 2x + y \ge 280 \quad \dots (1) \]
2) Phosphoric acid requirement (at least 14 kg):
\[ 6\% \text{ of } x + 10\% \text{ of } y \ge 14 \implies \frac{6x}{100} + \frac{10y}{100} \ge 14 \implies 3x + 5y \ge 700 \quad \dots (2) \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we obtain the unbounded feasible region shown below:

We now evaluate the objective function \( Z = 6x + 5y \) at the corner points of this unbounded region:

Since the feasible region is unbounded, we must check whether the half-plane \( 6x + 5y < 1000 \) has any points in common with the feasible region. Plotting this inequality, we find no common points except at the vertex \( B(100, 80) \).
Therefore, the farmer should buy 100 kg of fertilizer \( F_1 \) and 80 kg of fertilizer \( F_2 \) to minimize the cost to Rs. 1000.
In simple words: To keep fertilizer costs as low as possible while still meeting the soil's nitrogen and phosphoric acid needs, the farmer should buy 100 kg of \( F_1 \) and 80 kg of \( F_2 \). This results in a minimum total cost of Rs. 1000.

Exam Tip: For LPP with an unbounded region, never assume the minimum value obtained is the final solution. You must check whether the open half-plane \( ax + by < Z_{\text{min}} \) has any common points with the feasible region to verify the minimum.

Question 9. A cooperative society of farmers has 50 hectors of land to grow two crops X and Y. the profit from crops X and Y per hector are estimated as Rs. 10,500 and Rs. 9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 liters and10 liters per hector. Further, no more than 800 liters of herbicide should be used in order to protect fish and wild life. How much land should be allocated to each crop so as to maximize the total profit ?
Answer:
Let \( x \) and \( y \) denote the hectares of land allocated to crop X and crop Y, respectively.
We formulate the linear programming problem as follows:
Maximize the objective function for total profit:
\[ Z = 10500x + 9000y = 1500(7x + 6y) \]
Subject to the following constraints:
1) Maximum area of land available is 50 hectares:
\[ x + y \le 50 \]
2) Maximum liquid herbicide available is 800 liters:
\[ 20x + 10y \le 800 \implies 2x + y \le 80 \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities on the graph, the bounded feasible region is represented by the shaded region with vertices \( O(0,0) \), \( A(40,0) \), \( B(30,20) \), and \( C(0,50) \):

We evaluate the total profit \( Z = 1500(7x + 6y) \) at each corner point of the feasible region:

The maximum profit is Rs. 4,95,000, which occurs at the corner point \( B(30, 20) \).
Therefore, the cooperative society should allocate 30 hectares of land to crop X and 20 hectares of land to crop Y to maximize the total profit to Rs. 4,95,000.
In simple words: To make the highest profit of Rs. 4,95,000 under the land and chemical limits, the farmers should plant crop X on 30 hectares and crop Y on 20 hectares of their land.

Exam Tip: High-valued coefficients in objective functions can be factored out (such as expressing \( 10500x + 9000y \) as \( 1500(7x + 6y) \)) to make the calculations simpler and quicker during exams.

Question 10. A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hrs of machine time and 3 hrs of craftsman time. While a cricket bat takes 3hr of machine time and 1 hrs of craftsman time. In a day 42 hrs of machine time and 24 hrs of craftsman time. What number of rackets and bats must be made if the factory is to work at full capacity ?
Answer:
Let \( x \) be the number of tennis rackets made and \( y \) be the number of cricket bats made. Based on the requirements, we can construct the following table:

Our goal is to maximize the objective function representing total profit:
\[ Z = 20x + 10y \]
Subject to the following constraints:
1) Machine time constraint (at most 42 hours):
\[ 1.5x + 3y \le 42 \]
2) Craftsman time constraint (at most 24 hours):
\[ 3x + y \le 24 \]
Additionally, we have the non-negativity constraints:
\[ x \ge 0, \quad y \ge 0 \]
Plotting these inequalities, we locate the feasible region bounded by the corner points \( O(0,0) \), \( A(0,14) \), \( E(4,12) \), and \( B(8,0) \):

We evaluate the profit function \( Z = 20x + 10y \) at each corner point of the feasible region:

Operating at full capacity means utilizing all available hours, which corresponds to the point where both equations are fully met as equalities. Solving \( 1.5x + 3y = 42 \) and \( 3x + y = 24 \) gives the unique solution \( x = 4 \) and \( y = 12 \).
Therefore, the factory should produce 4 tennis rackets and 12 cricket bats to work at full capacity, achieving the maximum daily profit of Rs. 200.
In simple words: To use all available machine and craftsman hours fully and make the most profit, the factory must produce exactly 4 tennis rackets and 12 cricket bats daily. This will bring in a maximum profit of Rs. 200.

Exam Tip: "Full capacity" mathematically means the point of intersection of the boundary constraints, as it refers to the scenario where all resources are completely exhausted. Verify the coordinates of this point algebraically to ensure absolute accuracy.