Maharashtra Board Class 12 Maths Part 2 Chapter 6 Miscellaneous 6 Solutions

12th Commerce Maths 2 Chapter 6 Miscellaneous Exercise 6 Answers Maharashtra Board

Linear Programming Class 12 Commerce Maths 2 Chapter 6 Miscellaneous Exercise 6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 6 Linear Programming Miscellaneous Exercise 6 Questions and Answers.

Std 12 Maths 2 Miscellaneous Exercise 6 Solutions Commerce Maths

(I) Choose The Correct Alternative.

Question 1.The value of the objective function is maximized under linear constraints.
(a) at the centre of feasible region
(b) at (0, 0)
(c) at any vertex of feasible region.
(d) The vertex which is at maximum distance from (0, 0).
Answer: (c) at any vertex of feasible region.
In simple words: The maximum value of an objective function in a Linear Programming Problem (LPP) is always found at one of the corner points (vertices) of the feasible region, not inside or at the origin unless the origin is a vertex.

🎯 Exam Tip: Remember that the optimal solution (maximum or minimum) for an LPP always occurs at a vertex of the feasible region, which is a key principle of the corner point method.

 

Question 2.Which of the following is correct?
(a) Every LPP has on optional solution
(b) Every LPP has unique optional solution
(c) If LPP has two optional solutions then it has infinitely many solutions
(d) The set of all feasible solutions LPP may not be a convex set.
Answer: (c) If LPP has two optional solutions then it has infinitely many solutions
In simple words: If an LPP has more than one optimal solution, it implies that the objective function attains the same optimal value along an entire edge connecting two optimal vertices, leading to infinitely many optimal solutions.

🎯 Exam Tip: Multiple optimal solutions in LPP occur when the objective function's contour line is parallel to one of the edges of the feasible region, encompassing all points on that edge as optimal.

 

Question 3.Objective function of LPP is
(a) a constraint
(b) a function to be maximized or minimized
(c) a relation between the decision variables
(d) a feasible region.
Answer: (b) a function to be maximized or minimized
In simple words: An objective function in Linear Programming is the mathematical expression that represents the quantity (like profit or cost) that we aim to either maximize or minimize.

🎯 Exam Tip: Understanding the objective function is crucial as it defines what you are trying to achieve (e.g., maximize profit, minimize cost) within the given constraints.

 

Question 4.The maximum value of z = 5x + 3y. subject to the constraints 3x + 5y = 15; 5x + 2y \( \le \) 10, x, y \( \ge \) 0 is
(a) 235
(b) \( \frac{235}{9} \)
(c) \( \frac{235}{19} \)
(d) \( \frac{235}{3} \)
Answer: (c) \( \frac{235}{19} \)
In simple words: To find the maximum value of z, you need to identify the feasible region defined by the given constraints, determine its corner points, and then evaluate the objective function z = 5x + 3y at each of these corner points. The highest value obtained will be the maximum.

🎯 Exam Tip: Always clearly graph the feasible region, identify all corner points accurately, and systematically evaluate the objective function at each point to avoid errors in finding the optimal value.

 

Question 5.The maximum value of z = 10x + 6y. subject to the constraints 3x + y \( \le \) 12, 2x + 5y \( \le \) 34, x \( \ge \) 0, y \( \ge \) 0 is.
(a) 56
(b) 65
(c) 55
(d) 66
Answer: (a) 56
In simple words: First, plot the inequalities to define the feasible region. Then, find the coordinates of all corner points of this region by solving the intersecting equations. Finally, substitute these coordinates into the objective function z = 10x + 6y to find the maximum value among them.

🎯 Exam Tip: Pay close attention to the intersection points of the constraint lines, as these often represent the critical corner points where the objective function is evaluated for optimality.

 

Question 6.The point at which the maximum value of z = x + y subject to the constraint x + 2y \( \le \) 70, 2x + y \( \le \) 15, x \( \ge \) 0, y \( \ge \) 0 is
(a) (36, 25)
(b) (20, 35)
(c) (35, 20)
(d) (40, 15)
Answer: (d) (40, 15)
In simple words: Graph the feasible region from the inequalities, identify the corner points, and test each point in the objective function z = x + y to find which one yields the highest value. The point (40, 15) is not possible as it violates \(2x+y \le 15\). The problem statement might have a typo, assuming `x+2y <= 70` and `2x+y <= 15` and `x,y >= 0`. However, the provided answer `(40,15)` doesn't fit the constraints of the question as written. With the given constraints, `2x+y <= 15` is a very tight constraint. Let's assume there's a typo in the question or the answer. If the question was for a different set of constraints that would give (40,15), it's a vertex. If the question is exactly as written, then `(40,15)` makes `2x+y = 2(40)+15 = 80+15 = 95` which is not `\le 15`. This indicates a potential discrepancy in the source material. Assuming a corrected question yields (40,15) as a valid vertex for *some* LPP.

🎯 Exam Tip: Always double-check if the proposed answer point satisfies all given constraints. A common mistake is to pick a point that maximizes the objective function but lies outside the feasible region.

 

Question 7.Of all the points of the feasible region the optimal value of z is obtained at a point
(a) Inside the feasible region
(b) at the boundary of the feasible region
(c) at vertex of feasible region
(d) on x -axis
Answer: (c) at vertex of feasible region
In simple words: The fundamental theorem of linear programming states that the optimal solution (maximum or minimum) for any LPP always occurs at one of the corner points, also known as vertices, of its feasible region.

🎯 Exam Tip: This is a core concept; always remember that you only need to evaluate the objective function at the vertices of the feasible region to find the optimal solution.

 

Question 8.Feasible region; the set of points which satisfy
(a) The objective function
(b) All of the given function
(c) Some of the given constraints
(d) Only non-negative constraints
Answer: (b) All of the given function
In simple words: The feasible region in an LPP is the graphical area containing all possible points that simultaneously satisfy every single constraint, including the non-negative restrictions for the variables.

🎯 Exam Tip: A point is only considered feasible if it satisfies *all* constraints. If even one constraint is violated, the point is outside the feasible region.

 

Question 9.Solution of LPP to minimize z = 2x + 3y subjected to x \( \ge \) 0, y \( \ge \) 0, 1 \( \le \) x + 2y \( \le \) 10 is
(a) x = 0, y = \( \frac{1}{2} \)
(b) x = \( \frac{1}{2} \), y = 0
(c) x = 1, y = -2
(d) x = y = \( \frac{1}{2} \)
Answer: (a) x = 0, y = \( \frac{1}{2} \)
In simple words: To minimize z, you need to find the feasible region defined by \(x \ge 0\), \(y \ge 0\), \(x + 2y \ge 1\), and \(x + 2y \le 10\). Then, evaluate z = 2x + 3y at the corner points of this region to find the minimum value. The point (0, 1/2) corresponds to a vertex.

🎯 Exam Tip: For minimization problems, ensure you are identifying the correct vertices and accurately calculating the objective function value at each to find the lowest possible result.

 

Question 10.The corner points of the feasible region given by the inequation x + y \( \le \) 4, 2x + y \( \le \) 7, x \( \ge \) 0, y \( \ge \) 0, are
(a) (0, 0), (4, 0), (3, 1), (0, 4)
(b) (0, 0), (\( \frac{7}{2} \), 0), (3, 1), (0, 4)
(c) (0, 0), (\( \frac{7}{2} \), 0), (3, 1), (5, 7)
(d) (6, 0), (4, 0), (3, 1), (0, 7)
Answer: (b) (0, 0), (\( \frac{7}{2} \), 0), (3, 1), (0, 4)
In simple words: The corner points are the vertices of the feasible region formed by the intersection of the given inequalities. You find them by solving pairs of linear equations that define the boundary lines of the region.

🎯 Exam Tip: Accurately identifying all corner points is fundamental to solving LPPs. Systematically list the intercepts and intersection points of the boundary lines, then verify if they lie within the feasible region.

 

Question 11.The corner point of the feasible region are (0, 0), (2, 0), (\( \frac{12}{3} \), \( \frac{3}{7} \)) and (0, 1) then the point of maximum z = 6.5x + y = 13
(a) (0, 0)
(b) (2, 0)
(c) (\( \frac{11}{3} \), \( \frac{3}{7} \))
(d) (0, 1)
Answer: (b) (2, 0)
In simple words: Substitute each given corner point into the objective function z = 6.5x + y and calculate the value of z. The point that yields the highest z-value is the point of maximum.

🎯 Exam Tip: This question tests your ability to evaluate the objective function at given vertices. Be careful with fractional coordinates and arithmetic.

 

Question 12.If the corner points of the feasible region are (0, 0), (3, 0), (2, 1) and (0, \( \frac{7}{3} \)) the maximum value of z = 4x + 5y is
(a) 12
(b) 13
(c) \( \frac{35}{2} \)
(d) 0
Answer: (b) 13
In simple words: Evaluate the objective function z = 4x + 5y at each given corner point: (0,0) gives 0; (3,0) gives 12; (2,1) gives 4(2)+5(1) = 8+5 = 13; (0, 7/3) gives 5(7/3) = 35/3 \( \approx \) 11.67. The maximum is 13 at (2,1).

🎯 Exam Tip: When given corner points, systematically evaluate the objective function for each point. Keep track of calculations, especially with fractions, to correctly identify the maximum or minimum value.

 

Question 13.If the comer points of the feasible region are (0, 10), (2, 2), and (4, 0) then the point of minimum z = 3x + 2y is.
(a) (2, 2)
(b) (0, 10)
(c) (4, 0)
(d) (2, 4)
Answer: (a) (2, 2)
In simple words: Calculate the value of z = 3x + 2y for each given corner point. For (0,10), z = 20. For (2,2), z = 3(2)+2(2) = 6+4 = 10. For (4,0), z = 3(4)+2(0) = 12. The minimum value of 10 occurs at (2,2).

🎯 Exam Tip: For minimization problems, ensure you correctly identify the lowest value after evaluating the objective function at all feasible corner points.

 

Question 14.The half plane represented by 3x + 2y \( \le \) 0 contains the point.
(a) (1, \( \frac{2}{2} \))
(b) (2, 1)
(c) (0, 0)
(d) (5, 1)
Answer: (c) (0, 0)
In simple words: A point is contained in the half-plane if, when its coordinates are substituted into the inequality, the inequality holds true. For 3x + 2y \( \le \) 0, only (0,0) satisfies it, as 3(0) + 2(0) = 0, which is \( \le \) 0. All other options would result in a positive value.

🎯 Exam Tip: To check if a point lies in a given half-plane, simply substitute its coordinates into the inequality. If the inequality holds true, the point is in the region.

 

Question 15.The half plane represented by 4x + 3y \( \ge \) 14 contains the point
(a) (0, 0)
(b) (2, 2)
(c) (3, 4)
(d) (1, 1)
Answer: (c) (3, 4)
In simple words: Substitute each option's coordinates into the inequality 4x + 3y \( \ge \) 14. For (0,0), 0 \( \not\ge \) 14. For (2,2), 4(2)+3(2) = 8+6 = 14, which is \( \ge \) 14. For (3,4), 4(3)+3(4) = 12+12 = 24, which is \( \ge \) 14. For (1,1), 4(1)+3(1) = 7, which is \( \not\ge \) 14. Both (2,2) and (3,4) satisfy, but typically these questions look for one best fit or a primary choice, or perhaps the problem implies a specific context for "contains the point". In the context of options given, (3,4) gives a value well above 14. If (2,2) is an option, it also satisfies. Assuming the provided answer (c) is correct, it means (3,4) is the intended point among the choices that contains the half plane.

🎯 Exam Tip: When multiple options satisfy an inequality, ensure you check all of them. If the question asks for *a* point, any valid option is technically correct. If it asks for *the* point, there might be additional context or a single unique answer expected from the specific setup.

 

(II) Fill In The Blanks.

 

Question 1.Graphical solution set of the in equations x \( \ge \) 0, y \( \ge \) 0 is in __________ quadrant.
Answer: First
In simple words: The region where both x and y coordinates are non-negative is defined as the first quadrant in a Cartesian coordinate system.

🎯 Exam Tip: Understanding the coordinate system's quadrants is fundamental for accurately sketching feasible regions in Linear Programming.

 

Question 2.The region represented by the in equations x \( \ge \) 0, y \( \ge \) 0 lines in __________ quadrants.
Answer: First
In simple words: The conditions x \( \ge \) 0 and y \( \ge \) 0 collectively define the area of the graph that lies in the first quadrant, where both variables are positive or zero.

🎯 Exam Tip: Non-negativity constraints are standard in LPPs, restricting the feasible region to the first quadrant, which simplifies the graphical analysis.

 

Question 3.The optimal value of the objective function is attained at the __________ points of feasible region.
Answer: End
In simple words: The optimal value (maximum or minimum) for an objective function in a Linear Programming problem is always found at the extreme points, or vertices, of the feasible region. "End points" refers to these vertices.

🎯 Exam Tip: This is a fundamental theorem in LPP; always focus on the corner points when seeking the optimal solution.

 

Question 4.The region represented by the inequality y \( \le \) 0 lies in __________ quadrants
Answer: Third and Fourth
In simple words: The inequality y \( \le \) 0 means that all points must have a y-coordinate that is zero or negative, which corresponds to the lower half of the Cartesian plane, encompassing the third and fourth quadrants.

🎯 Exam Tip: Clearly understanding how inequalities translate to regions on a graph is essential for accurately plotting the feasible region.

 

Question 5.The constraint that a factory has to employ more women (y) than men (x) is given by __________
Answer: y > x
In simple words: If the number of women (y) must be strictly greater than the number of men (x), the mathematical inequality representing this constraint is y > x.

🎯 Exam Tip: Correctly translating word problems into mathematical inequalities is a critical first step in formulating LPPs. Pay attention to strict inequalities vs. less than or equal to/greater than or equal to.

 

Question 6.A garage employs eight men to work in its showroom and repair shop. The constants that there must be not least 3 men in showroom and repair shop. The constrains that there must be at least 3 men in showroom and at least 2 men in repair shop are __________ and __________ respectively.
Answer: x \( \ge \) 3 and y \( \ge \) 2
In simple words: If 'x' represents men in the showroom and 'y' men in the repair shop, "at least 3 men in showroom" translates to x \( \ge \) 3, and "at least 2 men in repair shop" translates to y \( \ge \) 2.

🎯 Exam Tip: "At least" always translates to "greater than or equal to" (\( \ge \)), while "at most" translates to "less than or equal to" (\( \le \)).

 

Question 7.A train carries at least twice as many first class passengers (y) as second class passangers (x). The constraint is given by __________
Answer: y \( \ge \) 2x
In simple words: If 'y' is the number of first-class passengers and 'x' is the number of second-class passengers, "at least twice as many first class passengers as second class" means y must be greater than or equal to two times x.

🎯 Exam Tip: When translating word problems, correctly identify which variable is being compared and the nature of the comparison (e.g., "at least," "at most," "exactly").

 

Question 8.A dishwashing machine hold up to 40 pieces of large crockery (x) this constraint is given by __________
Answer: x \( \le \) 40
In simple words: If 'x' is the number of crockery pieces, "hold up to 40" means the machine can hold 40 pieces or fewer, which is expressed as x \( \le \) 40.

🎯 Exam Tip: "Hold up to" or "maximum of" translates to "less than or equal to" (\( \le \)), indicating an upper limit.

 

(III) State Whether Each Of The Following Is True Or False.

 

Question 1.The region represented by the inequalities x \( \ge \) 0, y \( \ge \) 0 lies in first quadrant.
Answer: True
In simple words: The first quadrant is defined by all points where both the x-coordinate and the y-coordinate are non-negative.

🎯 Exam Tip: This is a fundamental concept of the Cartesian coordinate system, essential for correctly plotting feasible regions.

 

Question 2.The region represented by the inqualities x \( \le \) 0, y \( \le \) 0 lies in first quadrant.
Answer: False
In simple words: The inequalities x \( \le \) 0 and y \( \le \) 0 define the third quadrant, where both x and y coordinates are negative or zero, not the first quadrant.

🎯 Exam Tip: Be precise with the signs of inequalities; they determine the exact region in the coordinate plane.

 

Question 3.The optimum value of the objective function of LPP occurs at the center of the feasible region.
Answer: False
In simple words: The optimal value of an LPP always occurs at one of the corner points (vertices) of the feasible region, not necessarily at its geometric center.

🎯 Exam Tip: It is a common misconception that the optimum is at the center; always recall the corner point theorem for LPPs.

 

Question 4.Graphical solution set of x \( \le \) 0, y \( \ge \) 0 in xy system lies in second quadrant.
Answer: True
In simple words: The second quadrant is characterized by points where the x-coordinate is negative or zero, and the y-coordinate is positive or zero.

🎯 Exam Tip: A clear understanding of coordinate system quadrants is vital for accurate graphical representation of inequalities.

 

Question 5.Saina wants to invest at most Rs. 24000 in bonds and fixed deposits. Mathematically this constraints is written as x + y \( \le \) 24000 where x is investment in bond and y is in fixed deposits.
Answer: True
In simple words: "At most Rs. 24000" means the total investment (x + y) must be less than or equal to 24000, which is correctly represented as x + y \( \le \) 24000.

🎯 Exam Tip: Correctly translating "at most" to \( \le \) is crucial for accurately formulating LPP constraints from word problems.

 

Question 6.The point (1, 2) is not a vertex of the feasible region bounded by 2x + 3y \( \le \) 6, 5x + 3y \( \le \) 15, x \( \ge \) 0, y \( \ge \) 0.
Answer: True
In simple words: To be a vertex, (1,2) must satisfy at least two of the boundary equations. Substituting (1,2) into 2x+3y=6 gives 2(1)+3(2)=8, which is not 6. Thus, (1,2) is not on this boundary line, and therefore cannot be a vertex formed by these constraints.

🎯 Exam Tip: A vertex is an intersection point of boundary lines. To verify if a point is a vertex, ensure it satisfies the equations of at least two intersecting constraint lines.

 

Question 7.The feasible solution of LPP belongs to only quadrant I. The Feasible region of graph x + y \( \le \) 1 and 2x + 2y \( \ge \) 6 exists.
Answer: True
In simple words: For x+y \( \le \) 1 and 2x+2y \( \ge \) 6 (which simplifies to x+y \( \ge \) 3), there is no region that satisfies both simultaneously (a value cannot be \( \le \) 1 and \( \ge \) 3 at the same time). Thus, the feasible region is empty, meaning it "exists" as an empty set. The first part, "feasible solution belongs to only quadrant I", is generally true due to non-negativity constraints in typical LPPs, but the statement as a whole refers to the existence of a region for the specific inequalities. In this case, no such region exists. There's a slight ambiguity in "exists". If it means "a non-empty region exists", then false. If it means "a defined region, possibly empty, exists", then true. Given "True" as the answer, it likely interprets "exists" as a valid definition, even if empty.

🎯 Exam Tip: Pay attention to contradictory constraints. If two constraints cannot be satisfied simultaneously, the feasible region is empty, and no optimal solution can be found.

 

(IV) Solve The Following Problems.

 

Question 1.Maximize z = 5x\( _1 \) + 6x\( _2 \), Subject to 2x\( _1 \) + 3x\( _2 \) \( \le \) 18, 2x\( _1 \) + X\( _2 \) \( \le \) 12, x \( \ge \) 0, y \( \ge \) 0
Solution:

Inequation	Corresponding equation	x\( _1 \)	x\( _2 \)	Points	Region
2x\( _1 \) + 3x\( _2 \) \( \le \) 18	2x\( _1 \) + 3x\( _2 \) = 18	0	6	A (0, 6)	Origin side
		9	0	B (9, 0)
2x\( _1 \) + x\( _2 \) \( \le \) 12	2x\( _1 \) + x\( _2 \) = 12	0	12	C (0, 12)	Origin side
		6	0	D (6, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक ग्राफ है जो रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र (feasible region) को दर्शाता है। इसमें x1 और x2 अक्षों पर दो असमानताएं (2x1 + 3x2 = 18 और 2x1 + x2 = 12) दर्शाई गई हैं। संभव क्षेत्र OADE एक बहुभुज है, जिसके कोने O(0,0), A(0,6), E(4.5,3) और D(6,0) हैं। E बिंदु दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
OAED is the feasible region O(0, 0) A(0, 6) D (6, 0) and E is the intersection of 2x\( _1 \) + 3x\( _2 \) = 18 and 2x\( _1 \) + x\( _2 \) = 12 For E, Solving, 2x\( _1 \) + 3x\( _2 \) = 18 .......(i) 2x\( _1 \) + x\( _2 \) = 12 ......(ii) We get x\( _1 \) = 4.5, x\( _2 \) = 3
\( \implies \) E = (4.5, 3)

Corner Points	Value of z = 5x\( _1 \) + 6x\( _2 \)
O (0, 0)	5\( \times \)0 + 6\( \times \)0 = 0
A (0, 6)	5\( \times \)0 + 6\( \times \)6 = 36
E (4.5, 3)	5 \( \times \) 4.5 + 6 \( \times \) 3 = 40.5
D (6, 0)	5\( \times \)6 + 6\( \times \)0 = 30

\( \therefore \) Maximum value of z = 40.5 at E(4.5, 3)
In simple words: We first define the feasible region using the given constraints by finding the intersection points and intercepts. Then, we evaluate the objective function z = 5x\( _1 \) + 6x\( _2 \) at each of these corner points. The highest value, 40.5, occurs at E(4.5, 3), which is the maximum profit.

🎯 Exam Tip: In maximization problems, always ensure that all corner points of the feasible region are correctly identified and evaluated to find the true maximum value.

 

Question 2.Minimize z = 4x + 2y, Subject to 3x + y \( \ge \) 27, x + y \( \ge \) 21, x \( \ge \) 0, y \( \ge \) 0
Solution:

Corner Points	Value of z = 4x + 2y
A (0, 27)	4\( \times \)0 + 2\( \times \)27 = 54
E (3, 18)	4\( \times \)3 + 2\( \times \)18 = 48
D (21, 0)	4\( \times \)21 + 2\( \times \) 0 = 84

AED is the feasible region A(0, 27), D(21, 0) and E is the point of intersection of 3x + y = 27 and x + y = 21 For E, Solving 3x + y = 27 .......(i) x + y = 21 .......(ii) We get x = 3, y = 18
\( \implies \) E(3, 18)

Inequation	Corresponding equation	x	y	Points	Region
3x + y \( \ge \) 27	3x + y = 27	0	27	A (0, 27)	Non-origin side
		9	0	B (9, 0)
x + y \( \ge \) 21	x + y = 21	0	21	C (0, 21)	Non-origin side
		21	0	D (21, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक रैखिक प्रोग्रामिंग समस्या के संभव क्षेत्र को दर्शाता है, जहाँ लक्ष्य z = 4x + 2y को न्यूनतम करना है। ग्राफ पर दो असमानताएं (3x + y = 27 और x + y = 21) और x, y \( \ge \) 0 की शर्तें हैं। संभव क्षेत्र (AED) असीमित है, जिसके कोने A(0,27), E(3,18) और D(21,0) हैं। E दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
\( \therefore \) Minimum value of z = 48 at (3, 18)
In simple words: We first graph the constraints to identify the feasible region, which is unbounded in this case. We then find the corner points of this region, including any intersection points of the boundary lines. By evaluating the objective function z = 4x + 2y at each corner point, we find the minimum value is 48, which occurs at E(3, 18).

🎯 Exam Tip: For unbounded feasible regions in minimization problems, always verify if a minimum actually exists by checking values beyond the corner points in the direction of unboundedness; however, for typical LPPs, the minimum will still be at a corner point if it exists.

 

Question 3.Maximize z = 6x + 10y, subject to 3x + 5y \( \le \) 10, 5x + 3y \( \le \) 15, x \( \ge \) 0, y \( \ge \) 0
Solution:

Inequation	Corresponding equation	x	y	Points	Region
3x + 5y \( \le \) 10	3x + 5y = 10	0	2	A (0, 2)	Origin side
		\( \frac{10}{3} \)	0	B (\( \frac{10}{3} \), 0)
5x + 3y \( \le \) 15	5x + 3y = 15	0	5	C (0, 5)	Origin side
		3	0	D (3, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक रैखिक प्रोग्रामिंग समस्या के संभव क्षेत्र (feasible region) को दर्शाता है। इसमें दो असमानताएं (3x + 5y = 10 और 5x + 3y = 15) और x, y \( \ge \) 0 की शर्तें हैं। संभव क्षेत्र OAED एक बहुभुज है, जिसके कोने O(0,0), A(0,2), E(\( \frac{45}{16} \), \( \frac{5}{16} \)) और D(3,0) हैं। E बिंदु दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
OAED is the feasible region; O(0, 0), A (0, 2) D (3, 0) and E is the point of intersection of 3x + 5y = 10 and 5x + 3y = 15 For E, Solving 3x + 5y = 10 5x + 3y = 15 We get, x = \( \frac{45}{16} \), y = \( \frac{5}{16} \)
\( \implies \) E(\( \frac{45}{16} \), \( \frac{5}{16} \))

Corner Points	Value of z = 6x + 10y
O (0, 0)	6\( \times \)0 + 10\( \times \)0 = 0
A (0, 2)	6\( \times \)0 + 10\( \times \)2 = 20
E(\( \frac{45}{16} \), \( \frac{5}{16} \))	6\( \times \)\( \frac{45}{16} \) + 10 \( \times \)\( \frac{5}{16} \) = 20
D (3, 0)	6\( \times \)3 + 10\( \times \) 0 = 18

Since the maximum value of z = 20 at two points i.e, at A(0, 2) and E(\( \frac{45}{16} \), \( \frac{5}{16} \)). z is maximum at all points on segment AE. Hence it has infinite number of solutions.
In simple words: The feasible region is defined by the given constraints. After finding the corner points O(0,0), A(0,2), E(\( \frac{45}{16} \), \( \frac{5}{16} \)), and D(3,0), we evaluate the objective function z = 6x + 10y. The maximum value of z is 20, which occurs at both A(0,2) and E(\( \frac{45}{16} \), \( \frac{5}{16} \)). This means all points on the line segment AE also yield the maximum value, resulting in infinitely many solutions.

🎯 Exam Tip: When the maximum (or minimum) value of the objective function occurs at two adjacent corner points, it implies that all points on the line segment connecting these two vertices also yield the same optimal value, leading to infinitely many solutions.

 

Question 4.Minimize z = 2x + 3y, Subject to x - y \( \le \) 1, x + y \( \ge \) 3, x \( \ge \) 0, y \( \ge \) 0
Solution:

Inequation	Corresponding equation	x	y	Points	Region
x - y \( \le \) 1	x - y = 1	0	-1	A (0,-1)	Origin side
		1	0	B (1, 0)
x + y \( \ge \) 3	x + y = 3	0	3	C (0, 3)	Non-origin side
		3	0	D (3, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक रैखिक प्रोग्रामिंग समस्या के संभव क्षेत्र को दर्शाता है, जहाँ z = 2x + 3y को न्यूनतम करना है। इसमें असमानताएं x - y = 1 और x + y = 3, साथ ही x \( \ge \) 0, y \( \ge \) 0 की शर्तें हैं। संभव क्षेत्र (shaded portion CE) एक असीमित क्षेत्र है, जिसके कोने C(0,3) और E(2,1) हैं। E बिंदु दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
Shaded portion CE is the feasible region Where C = (0, 3) and E is the point of intersection of x - y = 1 and x + y = 3 For E, Solving x - y = 1 .......(i) x + y = 3 .........(ii) We get, x = 2, y = 1
\( \implies \) E(2, 1)

Corner Points	Value of z = 2x + 3y
C (0, 3)	2\( \times \)0 + 3\( \times \)3 = 9
E (2, 1)	2\( \times \)2 + 3\( \times \)1 = 7

\( \therefore \) Minimum value z = 7 at E(2, 1)
In simple words: We graph the inequalities x - y \( \le \) 1, x + y \( \ge \) 3, x \( \ge \) 0, y \( \ge \) 0 to find the feasible region. The corner points of this region are C(0, 3) and E(2, 1). Evaluating the objective function z = 2x + 3y at these points, we find that the minimum value of z is 7, which occurs at the point E(2, 1).

🎯 Exam Tip: When dealing with inequalities like x - y \( \le \) 1, remember that the origin (0,0) is a good test point to determine which side of the line to shade. For unbounded regions, always check the direction of improvement for the objective function.

 

Question 5.Maximize z = 4x\( _1 \) + 3x\( _2 \), Subject to 3x\( _1 \) + x\( _2 \) \( \le \) 15, 3x\( _1 \) + 4x\( _2 \) \( \le \) 24, x \( \ge \) 0, y \( \ge \) 0
Solution:

Inequation	Corresponding equation	x\( _1 \)	x\( _2 \)	Points	Region
3x\( _1 \) + x\( _2 \) \( \le \) 15	3x\( _1 \) + x\( _2 \) = 15	0	15	A (0, 15)	Origin side
		5	0	B (5, 0)
3x\( _1 \) + 4x\( _2 \) \( \le \) 24	3x\( _1 \) + 4x\( _2 \) = 24	0	6	C (0, 6)	Origin side
		8	0	D (8, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक रैखिक प्रोग्रामिंग समस्या के संभव क्षेत्र (feasible region) को दर्शाता है। लक्ष्य z = 4x1 + 3x2 को अधिकतम करना है। ग्राफ पर दो असमानताएं (3x1 + x2 = 15 और 3x1 + 4x2 = 24) और x1, x2 \( \ge \) 0 की शर्तें हैं। संभव क्षेत्र OCEB एक बहुभुज है, जिसके कोने O(0,0), C(0,6), E(4,3) और B(5,0) हैं। E बिंदु दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
OCEB is the feasible region where O(0, 0), C(0, 6), B(5, 0) E is the point of intersection of 3x\( _1 \) + x\( _2 \) = 15 and 3x\( _1 \) + 4x\( _2 \) = 24 For E, Solving 3x\( _1 \) + x\( _2 \) = 15 ........(i) 3x\( _1 \) + 4x\( _2 \) = 24 .......(ii) We get, x\( _1 \) = 4, x\( _2 \) = 3
\( \implies \) E(4, 3)

Corner Points	Value of z = 4x\( _1 \) + 3x\( _2 \)
O (0, 0)	4\( \times \)0 + 3\( \times \)0 = 0
C (0, 6)	4\( \times \)0 + 3\( \times \)6 = 18
B (5, 0)	4\( \times \)5 + 3\( \times \)0 = 20
E (4, 3)	4\( \times \)4 + 3\( \times \)3 = 25

\( \therefore \) Maximum value of z = 25 at (4, 3)
In simple words: The feasible region is determined by plotting the given constraints. The corner points are O(0,0), C(0,6), B(5,0), and E(4,3), where E is the intersection of the two main constraint lines. Evaluating the objective function z = 4x\( _1 \) + 3x\( _2 \) at these points, the maximum value obtained is 25 at E(4, 3).

🎯 Exam Tip: Accurately solving systems of linear equations to find intersection points is vital for identifying all corner points of the feasible region, which are critical for determining the optimal solution.

 

Question 6.Maximize z = 60x + 50y, Subject to x + 2y \( \le \) 40, 3x + 2y \( \le \) 60, x \( \ge \) 0, y \( \ge \) 0
Solution:

Inequation	Corresponding equation	x	y	Points	Region
x + 2y \( \le \) 40	x + 2y = 40	0	20	A (0, 20)	Origin side
		40	0	B (40, 0)
3x + 2y \( \le \) 60	3x + 2y = 60	0	30	C (0, 30)	Origin side
		20	0	D (20, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ एक रैखिक प्रोग्रामिंग समस्या के संभव क्षेत्र (feasible region) को दर्शाता है। लक्ष्य z = 60x + 50y को अधिकतम करना है। ग्राफ पर दो असमानताएं (x + 2y = 40 और 3x + 2y = 60) और x, y \( \ge \) 0 की शर्तें हैं। संभव क्षेत्र OADE एक बहुभुज है, जिसके कोने O(0,0), A(0,20), E(10,15) और D(20,0) हैं। E बिंदु दोनों रेखाओं का प्रतिच्छेदन बिंदु है।
OAED is the feasible region O(0, 0), A(0, 20), D(20, 0) E is x + 2y = 40 and 3x + 2y = 60 For E, Solving x + 2y = 40 3x + 2y = 60 We get, x = 10, y = 15
\( \implies \) E = (10, 15)

Corner Points	Value of z = 60x + 50y
O (0, 0)	60\( \times \)0 + 50 \( \times \) 0 = 0
A (0, 20)	60\( \times \)0 + 50 \( \times \) 20 = 1000
D (20, 0)	60 \( \times \) 20 + 50 \( \times \) 0 = 1200
E (10, 15)	60 \( \times \) 10 + 50 \( \times \) 15 = 1350

\( \therefore \) Maximum value of z = 1350 at E(10, 15).
In simple words: The feasible region is found by plotting the given constraints and the non-negativity conditions. The corner points are O(0,0), A(0,20), D(20,0), and E(10,15). Evaluating the objective function z = 60x + 50y at these points, the maximum value is 1350, which occurs at the intersection point E(10, 15).

🎯 Exam Tip: Always draw the graph carefully to identify the exact feasible region. For maximization problems, the optimal solution is typically at the farthest vertex from the origin (in the direction of increasing z).

 

Question 6.Maximize \( z = 60x + 50y \), Subject to \( x + 2y \le 40 \), \( 3x + 2y \le 60 \), \( x \ge 0, y \ge 0 \)
Solution:

Inequation	Corresponding equation	x	y	Points	Region
\( x + 2y \le 40 \)	\( x + 2y = 40 \)	0	20	A (0, 20)	Origin side
		40	0	B (40, 0)
\( 3x + 2y \le 60 \)	\( 3x + 2y = 60 \)	0	30	C (0, 30)	Origin side
		20	0	D (20, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र को दर्शाता है। x और y अक्ष चर का प्रतिनिधित्व करते हैं। दो बाधा रेखाएँ, \( x + 2y = 40 \) और \( 3x + 2y = 60 \), आलेखित की गई हैं। छायांकित क्षेत्र OAED संभव क्षेत्र को दर्शाता है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - O(0,0), A(0,20), E(10,15), और D(20,0) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

OAED is the feasible region O(0, 0), A(0, 20), D(20, 0)
E is \( x + 2y = 40 \) and \( 3x + 2y = 60 \)
For E, Solving \( x + 2y = 40 \)
\( 3x + 2y = 60 \)
We get, \( x = 10, y = 15 \)
\( \implies \) E = (10, 15)

Corner Points	Value of \( z = 60x + 50y \)
O (0, 0)	\( 60 \times 0 + 50 \times 0 = 0 \)
A (0, 20)	\( 60 \times 0 + 50 \times 20 = 1000 \)
D (20, 0)	\( 60 \times 20 + 50 \times 0 = 1200 \)
E (10, 15)	\( 60 \times 10 + 50 \times 15 = 1350 \)

\( \therefore \) Maximum value of \( z = 1350 \) at E(10, 15).
In simple words: The maximum profit for the given objective function and constraints is Rs. 1350, achieved when \( x=10 \) and \( y=15 \), which corresponds to point E.

🎯 Exam Tip: Always identify the corner points of the feasible region and evaluate the objective function at each point to find the optimal solution (maximum or minimum). Plotting the feasible region accurately is crucial for success.

 

Question 7.Minimize \( z = 4x + 2y \), Subject to \( 3x + y \ge 27 \), \( x + y \ge 21 \), \( x + 2y \ge 30 \), \( x \ge 0, y \ge 0 \)
Solution:

Inequation	Corresponding equation	x	y	Points	Region
\( 3x + y \ge 27 \)	\( 3x + y = 27 \)	0	27	A (0, 27)	Non-origin side
		9	0	B (9, 0)
\( x + y \ge 21 \)	\( x + y = 21 \)	0	21	C (0, 21)	Non-origin side
		21	0	D (21, 0)
\( x + 2y \ge 30 \)	\( x + 2y = 30 \)	0	15	E (0, 15)	Non-origin side
		30	0	F (30, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र AGHF को दर्शाता है। x और y अक्ष चर का प्रतिनिधित्व करते हैं। तीन बाधा रेखाएँ, \( 3x + y = 27 \), \( x + y = 21 \), और \( x + 2y = 30 \), आलेखित की गई हैं। छायांकित क्षेत्र AGHF संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - A(0,27), G(3,8), H(12,9), और F(30,0) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

AGHF is the feasible region where A(0, 27) F(30, 0)
G is the point of intersection of \( 3x + y = 27 \) and \( x + y = 21 \)
H is the point of intersection of \( x + y = 21 \) and \( x + 2y = 30 \)
For G, Solving \( 3x + y = 27 \) .......(i)
\( x + y = 21 \) .........(ii)
We get, \( x = 3, y = 18 \)
\( \implies \) G(3, 18)
For H, Solving \( x + y = 21 \) .......(i)
\( x + 2y = 30 \) ........(ii)
We get, \( x = 12, y = 9 \)
\( \implies \) H = (12, 9)

Corner Points	Value of \( z = 4x + 2y \)
A (0, 27)	\( 4 \times 0 + 2 \times 27 = 54 \)
G (3, 18)	\( 4 \times 3 + 2 \times 18 = 48 \)
H (12, 9)	\( 4 \times 12 + 2 \times 9 = 66 \)
F (30, 0)	\( 4 \times 30 + 2 \times 0 = 120 \)

\( \therefore \) Minimum value of \( z = 48 \) at G(3, 18).
In simple words: The minimum value of the objective function \( z = 4x + 2y \) is 48, which occurs at the corner point G(3, 18) of the feasible region.

🎯 Exam Tip: For minimization problems, the optimal solution is often found at a vertex of the feasible region closest to the origin, especially if the feasible region is unbounded and opens away from the origin. Always check all corner points.

 

Question 8.A carpenter makes chairs and table profit are Rs. 140 per chair and Rs. 210 per table Both products are processed on three machines, Assembling, Finishing and Polishing the time required for each product in hours and availability of each machine is given by the following table.

Product/Machines	Chair (x)	Table (y)	Available time (hours)
Assembling	3	3	36
Finishing	5	2	50
Polishing	2	6	60

Formulate and solve the following Linear programming problem using graphical method.
Solution:Let x be the no. of chair and y be the no. of table.
\( \therefore x \ge 0, y \ge 0 \)
Total profit = \( 140x + 210y \)
According to the table, the constraints can be written as
\( 3x + 3y \le 36 \)
\( 5x + 2y \le 50 \)
\( 2x + 6y \le 60 \)
\( \therefore \) The given LPP can be formulated as.
Maximize \( z = 140x + 210y \)
Subject to \( 3x + 3y \le 36 \), \( 5x + 2y \le 50 \), \( 2x + 6y \le 60 \), \( x \ge 0, y \ge 0 \).

Inequation	Corresponding equation	x	y	Points	Region
\( 3x + 3y \le 36 \)	\( 3x + 3y = 36 \)	0	12	A (0, 12)	Origin side
		12	0	B (12, 0)
\( 5x + 2y \le 52 \)	\( 5x + 2y = 52 \)	0	26	C (0, 26)	Origin side
		10.4	0	D (10.4, 0)
\( 2x + 6y \le 60 \)	\( 2x + 6y = 60 \)	0	10	E (0, 10)	Origin side
		30	0	F (30, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र OEGH को दर्शाता है। x और y अक्ष चर का प्रतिनिधित्व करते हैं। तीन बाधा रेखाएँ, \( 3x + 3y = 36 \), \( 5x + 2y = 52 \), और \( 2x + 6y = 60 \), आलेखित की गई हैं। छायांकित क्षेत्र OEGH संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - O(0,0), E(0,10), G(3,9), H(28/3, 8/3), और D(10.4,0) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

OEGH is the feasible region
O(0, 0), D(10.4, 0), E(0, 10)
For G, Solving \( 3x + 3y = 36 \) ........(i)
\( 2x + 6y = 60 \) .......(ii)
\( \implies \) G = (3, 9)
For H, Solving \( 5x + 2y = 52 \) .........(i)
\( 3x + 3y = 36 \) ........(ii)
We get, \( x = \frac{28}{3}, y = \frac{8}{3} \)
\( \implies \) H\( (\frac{28}{3}, \frac{8}{3}) \)

Corner Points	Value of \( z = 140x + 210y \)
O (0,0)	\( 140 \times 0 + 210 \times 0 = 0 \)
E (0, 10)	\( 140 \times 0 + 210 \times 10 = 2100 \)
G (3, 9)	\( 140 \times 3 + 210 \times 9 = 2310 \)
H\( (\frac{28}{3}, \frac{8}{3}) \)	\( 140 \times \frac{28}{3} + 210 \times \frac{8}{3} = 1306.67 + 560 = 1866.67 \)
D (10.4, 0)	\( 140 \times 10.4 + 210 \times 0 = 1456 \)

\( \therefore \) \( z \) is maximum at (3, 9) and maximum profit = Rs. 2310.
In simple words: The maximum profit of Rs. 2310 is achieved when the carpenter produces 3 chairs and 9 tables, as determined by evaluating the objective function at the corner points of the feasible region.

🎯 Exam Tip: Always present the final answer in the context of the problem, stating the quantity of each item to be produced and the resulting maximum (or minimum) value of the objective function.

 

Question 9.A company manufactures bicycles and tricycles, each of which must be processed through two machines A and B Maximum availability of machines A and B are respectively 120 and 180 hours. Manufacturing a bicycle requires 6 hours on machine A and 3 hours on machine B manufacturing a tricycle requires 4 hours on machine A and 10 hours on machine B. If profits are Rs. 180 for a bicycle and Rs. 220 on a tricycle, determine the number of bicycles and tricycles that should be manufactured in order to maximize the profit.
Solution:Let x number of bicycle and y number of tricycle has to be manufactured and to be sold to get the profit (z)
\( \therefore x \ge 0, y \ge 0 \)
Total profit = \( 180x + 220y \).
The given LPP can be tabulated as follows:

Machines	Bicycle	Tricycle	Available line
A (Hrs)	6	4	120
B (Hrs)	3	10	180

\( \therefore \) The given LPP can be formulated as
Maximize \( z = 180x + 220y \)
Subject to \( 6x + 4y \le 120 \), \( 3x + 10y \le 180 \), \( x \ge 0, y \ge 0 \).

Inequation	Corresponding equation	x	y	Points	Region
\( 6x + 4y \le 120 \)	\( 6x + 4y = 120 \)	0	30	A (0, 30)	Origin side
		20	0	B (20, 0)
\( 3x + 10y \le 180 \)	\( 3x + 10y = 180 \)	0	18	C (0, 18)	Origin side
		60	0	D (60, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र OCEB को दर्शाता है। x और y अक्ष क्रमशः साइकिल और तिपहिया वाहनों की संख्या का प्रतिनिधित्व करते हैं। दो बाधा रेखाएँ, \( 6x + 4y = 120 \) और \( 3x + 10y = 180 \), आलेखित की गई हैं। छायांकित क्षेत्र OCEB संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - O(0,0), C(0,18), E(10,15) और B(20,0) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

OCEB is the feasible region where O(0, 0) C(0, 18) B(20, 0)
E is the point of intersection of \( 6x + 4y = 120 \) and \( 3x + 10y = 180 \)
For E, Solving \( 6x + 4y = 120 \) ........(i)
\( 3x + 10y = 180 \) .........(ii)
We get, \( x = 10, y = 15 \)
\( \implies \) E(10, 15)

Corner Points	Value of \( z = 180x + 220y \)
O (0, 0)	\( 180 \times 0 + 220 \times 0 = 0 \)
C (0, 18)	\( 180 \times 0 + 220 \times 18 = 3960 \)
B (20, 0)	\( 180 \times 20 + 220 \times 0 = 3600 \)
E (10, 15)	\( 180 \times 10 + 220 \times 15 = 5100 \)

\( \therefore \) Maximum value of \( z \) is 5100 at E(10, 15)
Hence 10 bicycles and 15 tricycles should be produced to get maximum profit.
In simple words: To maximize profit, the company should produce 10 bicycles and 15 tricycles, which yields a maximum profit of Rs. 5100, identified at the intersection point E(10, 15).

🎯 Exam Tip: When formulating an LPP, clearly define variables, the objective function, and all constraints from the problem statement. Errors in formulation lead to incorrect feasible regions and optimal solutions.

 

Question 10.A factory produced two types of chemicals A and B. The following table gives the units of ingredients P and Q (per kg) of chemicals A and B as well as minimum requirements of P and Q and also cost per kg. Chemicals A and B:

Chemicals Units/Ingredients per kg.	A (x)	B (y)	Minimum requirements in
P	1	2	80
Q	3	1	75
Cost (in Rs.)	4	6

Find the number of units of chemicals A and B should be produced sp as to minimize the cost.
Solution:Let x be the no. of units of chemicals, A produced and y be the no. of units of chemical B produced.
Total cost is \( 4x + 6y \)
The LPP is. Minimise \( z = 4x + 6y \)
Subject to \( x + 2y \ge 80 \), \( 3x + y \ge 75 \), \( x \ge 0, y \ge 0 \).

Inequation	Corresponding equation	x	y	Points	Region
\( x + 2y \ge 80 \)	\( x + 2y = 80 \)	0	40	A (0, 40)	Non-origin side
		80	0	B (80, 0)
\( 3x + y \ge 75 \)	\( 3x + y = 75 \)	0	75	C (0, 75)	Non-origin side
		25	0	D (25, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र BEC को दर्शाता है। x और y अक्ष क्रमशः रसायन A और रसायन B की इकाइयों की संख्या का प्रतिनिधित्व करते हैं। दो बाधा रेखाएँ, \( x + 2y = 80 \) और \( 3x + y = 75 \), आलेखित की गई हैं। छायांकित क्षेत्र BEC संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - B(80,0), E(14,33), और C(0,75) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

The shaded region BEC in the feasible region B(80, 0) C(0, 75)
and E is the point of intersection of \( x + 2y = 80 \) and \( 3x + y = 75 \)
For E, Solving \( x + 2y = 80 \) .......(i)
\( 3x + y = 75 \) .......(ii)
We get, \( x = 14, y = 33 \)
\( \implies \) E(14, 33)

Corner Points	Value of \( z = 4x + 6y \)
B (80, 0)	\( 4 \times 80 + 6 \times 0 = 320 \)
E (14,33)	\( 4 \times 14 + 6 \times 33 = 254 \)
C (0, 75)	\( 4 \times 0 + 6 \times 75 = 450 \)

\( \therefore \) \( z \) is minimum at E(14, 33) and the minimum value of \( z = 254 \).
Hence 14 units of chemical A and 33 units of chemical B are produced to get a minimum cost of Rs. 254.
In simple words: To minimize the cost, 14 units of chemical A and 33 units of chemical B should be produced, resulting in a minimum cost of Rs. 254.

🎯 Exam Tip: For minimization problems, ensure the feasible region is correctly identified. The optimal solution will be one of the corner points. Double-check calculations for intersection points, as these are critical for accuracy.

 

Question 11.A company produces mixers and food processors. Profit on selling one mixer and one food processor is Rs. 2,000/- and Rs. 3,000/- respectively. Both the products are processed through three Machines A, B, C. The time required in hours by each product and total time available in hours per week on each machine are as follows:

Product Machine	Mixer per unit	Food processor per unit	Available time
A	3	3	36
B	5	2	50
C	2	6	60

How many mixes and food processors should be produced to maximize profit?
Solution:Let x be the no. of mixers produced and y be the no. of food processors produced.
The profit is \( 2000x + 3000y \)
The LPP is Maximize \( z = 2000x + 3000y \)
Subject to \( 3x + 3y \le 36 \), \( 5x + 2y \le 50 \), \( 2x + 6y \le 60 \), \( x \ge 0, y \ge 0 \).
Let \( 3x + 3y = 36 \),
\( \implies x + y = 12 \)
\( x = 0, y = 12, (0, 12) \)
\( y = 0, x = 12, (12, 0) \)
\( 5x + 2y = 50 \)
\( x = 0, y = 25, (0, 25) \)
\( y = 0, x = 10, (10, 0) \)
Let \( 2x + 6y = 60 \), \( x + 3y = 30 \)
\( x = 0, y = 10 (0, 10) \)
\( y = 0, x = 30 (30, 0) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र OABCD को दर्शाता है। x और y अक्ष क्रमशः मिक्सर और फूड प्रोसेसर की संख्या का प्रतिनिधित्व करते हैं। तीन बाधा रेखाएँ, \( x + y = 12 \), \( 5x + 2y = 50 \), और \( x + 3y = 30 \), आलेखित की गई हैं। छायांकित क्षेत्र OABCD संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - O(0,0), A(10,0), B(8.67,3.33), C(3,9) और D(0,10) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

The Shaded region OABCD is the feasible region.
O(0, 0) A (10, 0) D(0, 10)
B is the intersection of \( x + y = 12 \) and \( 5x + 2y = 50 \)
Solving we get \( x = 8.67, y = 3.33 \)
\( \implies \) B(8.67, 3.33)
C is the intersection of \( x + y = 12 \) and \( x + 3y = 30 \)
Solving we get \( x = 3, y = 9 \)
\( \implies \) C(3, 9)

Corner Points	Value of \( z = 2000x + 3000y \)
O (0, 0)	0
A (10, 0)	20,000
B (8.67, 3.33)	\( 17340 + 9990 = 27330 \)
C (3, 9)	\( 6000 + 27000 = 33000 \)
D (0, 10)	30000

\( \therefore \) Maximum value of \( z \) is 33000 at C(3, 9)
Hence 3 mixers and 9 food processors should be produced to get a maximum profit of Rs. 33,000.
In simple words: To achieve the maximum profit of Rs. 33,000, the company should produce 3 mixers and 9 food processors, as identified from the corner point C(3, 9) of the feasible region.

🎯 Exam Tip: When dealing with decimal values for intersection points, ensure consistent rounding (e.g., two decimal places) throughout the calculation. The optimal solution is typically at an integer vertex if the variables represent discrete units, but LPP often provides non-integer solutions which are then interpreted.

 

Question 12.A chemical company produces a chemical containing three basic elements A, B, C so that it has at least 16 liters of A, 24 liters of B, and 18 liters of C. This chemical is made by mixing two compounds I and II. Each unit of compound I has 4 liters of A, 12 liters of B, 2 liters of C. Each unit of compound II has 2 liters of A, 2 liters of B and 6 liters of C. The cost per unit of compound is Rs. 800/- and that of compound II is Rs. 640/- Formulate the problem as L.P.P and solve it to minimize the cost.
Solution:Let x be the no. of units of compound I used and y be the no of units of compound II used.
The data can be tabulated as

	A	B	C	Cost
I (x)	4	12	2	800
II (y)	2	2	6	640
Min. Requirement	16	24	18

The LPP is minimize \( z = 800x + 640y \)
Subject to \( 4x + 2y \ge 16 \), \( 12x + 2y \ge 24 \), \( 2x + 6y \ge 18 \), \( x, y \ge 0 \).
Let \( 4x + 2y = 16 \),
\( \implies 2x + y = 8 \)
\( x = 0, y = 8, (0, 8) \)
\( y = 0, x = 4, (4, 0) \)
\( 12x + 2y = 24 \),
\( \implies 6x + y = 12 \)
\( x = 0, y = 12, (0, 12) \)
\( y = 0, x = 2, (2, 0) \)
\( 2x + 6y = 18 \)
\( \implies x + 3y = 9 \)
\( x = 0, y = 3, (0, 3) \)
\( y = 0, x = 9, (9, 0) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र ABCD को दर्शाता है। x और y अक्ष क्रमशः यौगिक I और यौगिक II की इकाइयों की संख्या का प्रतिनिधित्व करते हैं। तीन बाधा रेखाएँ, \( 2x + y = 8 \), \( 6x + y = 12 \), और \( x + 3y = 9 \), आलेखित की गई हैं। छायांकित क्षेत्र ABCD संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - A(0,12), B(1,6), C(3,2) और D(9,0) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

The Shaded region ABCD is the feasible region.
A(0, 12) D(9, 0)
B is the intersection of \( 6x + y = 12 \) and \( 2x + y = 8 \)
Solving we get \( x = 1, y = 6 \)
\( \implies \) B(1, 6)
C is the intersection of \( 2x + y = 8 \) and \( x + 3y = 9 \)
Solving we get \( x = 3, y = 2 \)
\( \implies \) C(3, 2)

Corner Points	Value of \( z = 800x + 640y \)
A (0, 12)	\( 0 + 7680 = 7680 \)
B (1, 6)	\( 800 + 3840 = 4640 \)
C (3, 2)	\( 2400 + 1280 = 3680 \)
D (9, 0)	\( 7200 + 0 = 7200 \)

\( \therefore \) The minimum value of \( z = 3680 \) at (3, 2)
Hence 3 unit of compound I and 2 units of compound II should be used to get the minimum cost of Rs. 3680.
In simple words: To minimize the cost to Rs. 3680, the company should use 3 units of compound I and 2 units of compound II.

🎯 Exam Tip: When multiple constraints are present, carefully identify all intersection points to ensure no feasible corner point is missed. For "at least" constraints (\( \ge \)), the feasible region will be away from the origin.

 

Question 13.A person who makes two types of gift items A and B requires the services of a cutter and a finisher. Gift item A requires 4 hours of the cutter's time and 2 hours of the finisher's time. B required 2 hours of the cutter's time and 4 hours of finisher's time. The cutter and finisher have 208 hours and 152 hours available times respectively every month. The profit of one gift item of type A is Rs. 75/- and on gift item B is Rs. 125/-. Assuming that the person can sell all the gift items produced, determine how many gift items of each type should he make every month to obtain the best returns?
Solution:Let x be the no. of gift A produced and y be the no. of gift B produced.
The data can be tabulated as

	Cutter Hrs.	Finisher Hrs.	Profit (Rs.)
A (x)	4	2	75
B (y)	2	4	125
Available	208	152

The LLP is maximize \( z = 75x + 125y \)
Subject to \( 4x + 2y \le 208 \), \( 2x + 4y \le 152 \), \( x, y \ge 0 \)
Let \( 4x + 2y = 208 \), \( \implies 2x + y = 104 \)
\( x = 0, y = 104; (0, 104) \), \( x = 0, y = 38; (0, 38) \)
\( y = 0, x = 52; (52, 0) \), \( y = 0, x = 76; (76, 0) \)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक रैखिक प्रोग्रामिंग समस्या के लिए संभव क्षेत्र OABC को दर्शाता है। x और y अक्ष क्रमशः उपहार आइटम A और B की संख्या का प्रतिनिधित्व करते हैं। दो बाधा रेखाएँ, \( 2x + y = 104 \) और \( x + 2y = 76 \), आलेखित की गई हैं। छायांकित क्षेत्र OABC संभव क्षेत्र है, जो दिए गए बाधाओं को संतुष्ट करने वाले सभी बिंदुओं का समूह है। इस क्षेत्र के कोने बिंदु - O(0,0), A(52,0), B(44,16), और C(0,38) - इष्टतम समाधान खोजने के लिए महत्वपूर्ण हैं।

The shaded region OABC is the feasible region.
O(0,0) A (52, 0) C(0, 38)
B is the intersection of \( 2x + y = 104 \) and \( x + 2y = 76 \)
Solving we get \( x = 44, y = 16 \)
\( \implies \) B(44, 16)

Corner Points	Value of \( z = 75x + 125y \)
O (0, 0)	0
A (52, 0)	3900
B (44, 16)	\( 3300 + 2000 = 5300 \)
C (0, 38)	\( 0 + 4750 = 4750 \)

\( \therefore \) Maximum value of \( z = 5300 \) at B(44, 16)
Hence he should produce 44 gifts of type A & 16 of type B to get a maximum profit of Rs. 5300.
In simple words: To maximize profit, the person should produce 44 gift items of type A and 16 gift items of type B, resulting in a maximum profit of Rs. 5300.

🎯 Exam Tip: Always verify that the chosen solution (corner point) falls within the feasible region defined by all constraints. This prevents selecting an infeasible point that might appear optimal in isolated calculations.

 

Question 14.A firm manufactures two products A and B on which profit is earned per unit Rs. 3/- and Rs. 4/- respectively. Each product is processed on two machines M\( _1 \) and M\( _2 \). Product A requires one minute of processing time on M\( _1 \) and two minutes of processing time on M\( _2 \). B requires one minute of processing time on M\( _1 \) and one minute processing time on M\( _2 \) Machine M\( _1 \) is available for use for 450 minutes while M\( _2 \) is available for 600 minutes during any working day. Final the number of units of products A and B to be manufactured to get the maximum profit.
Solution:Let x denote the number of units of product A and y denote the number of units of product B.
The total profit in \( 3x + 4y \),
The given statements can be tabulated as

	A (x)	B (y)	Availability
M\( _1 \)	1	1	450
M\( _2 \)	2	1	600

The constraints are \( x + y \le 450 \), \( 2x + y \le 600 \), \( x \ge 0, y \ge 0 \)
\( \therefore \) The LPP can be formulated as
Maximize \( z = 3x + 4y \)
Subject to \( x + y \le 450 \), \( 2x + y \le 600 \), \( x \ge 0, y \ge 0 \).

Inequation	Corresponding equation	x	y	Points	Region
\( x + y \le 450 \)	\( x + y = 450 \)	0	450	A (0, 450)	Origin side
		450	0	B (450, 0)
\( 2x + y \le 600 \)	\( 2x + y = 600 \)	0	600	C (0,600)	Origin side
		300	0	D (300,0)

 

Question 15. A firm manufacturing two types of electrical items A and B, can make a profit of Rs. 20/- per unit of A and Rs. 30/- per unit of B. Both A and B make use of two essential components a motor and a transformer. Each unit of A requires 3 motors and 2 transformers and each unit of B required 2 motors and 4 transformers. The total supply of components per month is restricted to 210 motors and 300 transformers. How many units of A and B should they manufacture per month to maximize profit? How much is the maximum profit?
Answer:
Solution:
Let x be the no. of electrical item A and y be the no. of electrical items to be manufactured per month to maximize the profit.
Total profit is Rs. 20x + 30y
The given condition can be tabulated as.

	A (x)	B (y)	Availability
Motor	3	2	210
Transformer	2	4	300

The given LPP can be formulated as
Maximize \(z = 20x + 30y\)
Subject to \(3x + 2y \leq 210\)
\(2x + 4y \leq 300\)
\(x \geq 0, y \geq 0\).

Inequation	Corresponding equation	x	y	Points	Region
\(3x+2y \leq 210\)	\(3x+2y = 210\)	0	105	A (0, 105)	Origin side
		70	0	B (70, 0)
\(2x+4y \leq 300\)	\(2x+4y = 300\)	0	75	C (0, 75)	Origin Side
		150	0	D (150, 0)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक ग्राफ है जो रैखिक प्रोग्रामिंग समस्या के व्यवहार्य क्षेत्र को दर्शाता है। क्षैतिज अक्ष (X) और ऊर्ध्वाधर अक्ष (Y) उत्पादन की इकाइयों का प्रतिनिधित्व करते हैं। नीले रंग का छायांकित क्षेत्र व्यवहार्य क्षेत्र (BOCE) है, जो सभी बाधाओं को संतुष्ट करने वाले बिंदुओं का सेट है। बिंदु B, O, C, और E इस व्यवहार्य क्षेत्र के शीर्ष हैं, जहां E दो मुख्य बाधा रेखाओं \(3x+2y = 210\) और \(2x+4y = 300\) का प्रतिच्छेदन बिंदु है।

BOCE is the feasible region
B(70, 0), O(0, 0), C(0, 75)
E is the intersection of \(3x + 2y = 210\) and \(2x + 4y = 300\)
For E, Solving \(3x + 2y = 210\) ..........(i)
\(2x + 4y = 300\) .......(ii)
We get \(x = 30, y = 60\)
Hence \(E(30, 60)\)

Corner Points	Value of \(z = 20x + 30y\)
B (70, 0)	\(20 \times 70 + 30 \times 0 = 1400\)
O (0, 0)	\(20 \times 0 + 30 \times 0 = 0\)
C (0, 75)	\(20 \times 0 + 30 \times 75 = 2250\)
E (30, 60)	\(20 \times 30 + 30 \times 60 = 2400\)

Therefore, the maximum value of \(z = \text{Rs. } 2400\) at (30, 60)
Hence 30 units of A and 60 units of B should be manufactured per month to get the maximum profit of Rs. 2400
In simple words: To achieve the highest profit, the company should produce 30 units of electrical item A and 60 units of electrical item B, resulting in a maximum profit of Rs. 2400. This is found by evaluating the objective function at the corner points of the feasible region defined by the production constraints.

🎯 Exam Tip: Always clearly define variables, set up the objective function and constraints correctly, and ensure all corner points of the feasible region are accurately calculated for evaluation.

12th Commerce Maths Digest Pdf

• 12th Commerce Maths Exercise 6.1 Solutions
• 12th Commerce Maths Exercise 6.2 Solutions
• 12th Commerce Maths Miscellaneous Exercise 6 Solutions