68 free MCQs on Linear Programming with worked answers and explanations. Optimizing a linear objective function subject to linear constraints using the graphical corner point method.
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The feasible region (shaded) is bounded by the constraint lines; the fundamental theorem of LPP guarantees the optimal value of the objective function occurs at one of the corner points (O, A, B, C), so only these need to be checked, not the entire region.
Q1.
In a linear programming problem, the function to be maximized or minimized is called the:
Answer: B. Objective function
Why: The objective function, usually written as Z = ax + by, is the linear function whose optimal value (maximum or minimum) we seek.
Q2.
The unknown quantities that a linear programming problem solves for are called:
Answer: B. Decision variables
Why: Decision variables (commonly x and y) represent the quantities to be determined, such as units of a product to manufacture.
Q3.
The linear inequalities that decision variables must satisfy in an LPP are called:
Answer: A. Constraints
Why: Constraints are the linear inequalities (and sometimes equations) representing limitations such as available resources, time, or material.
Q4.
The set of all points satisfying every constraint of an LPP simultaneously is called the:
Answer: B. Feasible region
Why: The feasible region is the common region (intersection) of all the constraint half-planes, containing every feasible solution.
Q5.
According to the fundamental theorem of linear programming, if an optimal value exists, where does it occur?
Answer: B. At a corner point of the feasible region
Why: If an LPP has an optimal solution, it must occur at one of the corner points (vertices) of the feasible region.
Q6.
A feasible region that can be enclosed within a circle of finite radius is called:
Answer: B. Bounded
Why: A bounded feasible region does not extend infinitely in any direction and can be enclosed in a circle of finite radius.
Q7.
Decision variables in a typical LPP are restricted to be:
Answer: B. Non-negative (x >= 0, y >= 0)
Why: Since decision variables usually represent real quantities like units produced, they are restricted to non-negative values.
Q8.
A diet problem in linear programming typically aims to:
Answer: B. Minimize cost while meeting nutritional requirements
Why: In a diet problem, the objective is usually to minimize the cost of a diet while satisfying minimum requirements of nutrients like proteins and vitamins.
Q9.
A manufacturing problem in linear programming typically aims to:
Answer: B. Maximize profit subject to limited resources
Why: A manufacturing problem usually maximizes profit (the objective function) subject to constraints like limited labour hours, machine hours, or raw material.
Q10.
The graphical method of solving an LPP by evaluating the objective function at every vertex of the feasible region is called the:
Answer: B. Corner point method
Why: The corner point method evaluates Z at each corner point of the feasible region to find the maximum or minimum value.
Q11.
Linear programming is a method for finding the ___ value of a linear function subject to constraints:
Answer: A. the optimal (best) value
Why: An LPP finds the optimal (maximum or minimum) value of a linear objective function.
Q12.
The linear function to be maximised or minimised in an LPP is called the:
Answer: A. objective function
Why: The objective function, e.g. Z = ax + by, is what we optimise.
Q13.
The conditions written as linear inequalities in an LPP are called:
Answer: A. constraints
Why: Constraints are the linear inequalities that limit the variables.
Q14.
The common region that satisfies all the constraints is the:
Answer: A. feasible region
Why: The feasible region is the set of all points meeting every constraint.
Q15.
The requirement that the variables be greater than or equal to zero is the ___ restriction:
Answer: A. non-negativity
Why: The non-negativity restrictions x ≥ 0, y ≥ 0 keep the solution in the first quadrant.
Q16.
The optimal solution of an LPP occurs at a ___ of the feasible region:
Answer: A. corner point
Why: The optimum lies at a corner (vertex) of the feasible region.
Q17.
A feasible region that extends indefinitely is described as:
Answer: A. unbounded
Why: An unbounded feasible region is not enclosed and stretches out without limit.
Q18.
With x ≥ 0 and y ≥ 0, the feasible region lies in the:
Answer: A. first quadrant
Why: Both coordinates non-negative places the region in the first quadrant.
Q19.
Points that satisfy all the constraints of an LPP are called ___ solutions:
Answer: A. feasible
Why: Any point in the feasible region is a feasible solution.
Q20.
A linear programming problem has a linear objective function together with linear:
Answer: A. constraints
Why: Both the objective and the constraints are linear in an LPP.
Q21.
Maximize Z = 3x + 4y subject to x + y <= 4, x >= 0, y >= 0. What is the maximum value of Z?
Answer: B. 16
Why: Corner points are (0,0), (4,0), (0,4). Z values: 0, 12, 16. Maximum Z = 16 at (0,4).
Q22.
For the feasible region with corners (0,0), (4,0), (0,4) from x + y <= 4 (x,y >= 0), what is the maximum value of Z = 5x + 2y?
Answer: A. 20
Why: Evaluate Z at each corner: Z(0,0)=0, Z(4,0)=20, Z(0,4)=8. Maximum Z = 20 at (4,0).
Q23.
A transportation problem in linear programming typically minimizes:
Answer: B. Total transportation cost while meeting supply and demand
Why: A transportation problem minimizes the total cost of moving goods from sources (factories or warehouses) to destinations (markets), subject to supply and demand constraints.
Q24.
Maximize Z = 50x + 60y subject to 2x + y <= 120, x + 2y <= 120, x >= 0, y >= 0. The corner points of the feasible region are (0,0), (60,0), (40,40), and (0,60). What is the maximum value of Z?
Answer: A. 4400
Why: Z(0,0)=0, Z(60,0)=3000, Z(40,40)=50(40)+60(40)=2000+2400=4400, Z(0,60)=3600. Maximum is 4400 at (40,40).
Q25.
If the feasible region of an LPP is unbounded, what is true about finding the maximum of the objective function?
Answer: B. A maximum may not exist even if the corner point values suggest one, and must be checked separately
Why: On an unbounded feasible region, a value that looks like the maximum from corner points must be verified by checking whether the half-plane Z > that value has any point in common with the feasible region; if it does, no maximum exists.
Q26.
Minimize Z = 200x + 500y subject to x + 2y >= 10, 3x + 4y <= 24, x >= 0, y >= 0. The feasible region has corners (0,5), (4,3), and (0,6). What is the minimum value of Z?
Answer: A. 2300
Why: Z(0,5)=2500, Z(4,3)=200(4)+500(3)=800+1500=2300, Z(0,6)=3000. Minimum Z = 2300 at (4,3).
Q27.
In an LPP, what does a corner point represent geometrically?
Answer: B. A point where two boundary constraint lines (or a boundary line and an axis) intersect
Why: Corner points (vertices) occur where two boundary lines of the feasible region meet, including intersections with the coordinate axes.
Q28.
Maximize Z = 7x + 4y subject to 3x + y <= 600, x + y <= 300, x >= 0, y >= 0. The relevant corner points are (0,0), (200,0), (150,150), and (0,300). What is the maximum Z?
Answer: A. 1650
Why: Z(0,0)=0, Z(200,0)=1400, Z(150,150)=7(150)+4(150)=1050+600=1650, Z(0,300)=1200. Maximum Z = 1650 at (150,150).
Q29.
If two corner points of a feasible region give the same maximum value of the objective function, what can be concluded?
Answer: B. Every point on the line segment joining those two corners also gives the same maximum value
Why: When the objective function takes the same optimal value at two corner points, it takes that same value at every point on the segment joining them, giving infinitely many optimal solutions.
Q30.
For constraints x >= 0, y >= 0, and x <= 10 only (no upper bound on y), the feasible region is:
Answer: B. Unbounded
Why: Since y has no upper bound, the feasible region extends infinitely upward, making it unbounded.
Q31.
By the corner-point theorem, the optimum of the objective function is attained at a:
Answer: A. vertex of the feasible region
Why: The maximum or minimum, if it exists, occurs at a vertex (corner point) of the feasible region.
Q32.
If the feasible region of an LPP is bounded, the objective function has:
Answer: A. both a maximum and minimum
Why: On a bounded feasible region the objective attains both a maximum and a minimum value.
Q33.
To solve an LPP graphically, the objective function is evaluated at each:
Answer: A. corner point
Why: We compute Z at every corner point and pick the best value.
Q34.
Maximise Z = 3x + 2y over the corner points (0,0), (4,0) and (0,5). The maximum value is:
Answer: A. 12
Why: Z(4,0) = 12, Z(0,5) = 10, Z(0,0) = 0; the maximum is 12.
Q35.
The constraints of an LPP usually take a form such as ax + by ≤ c or:
Answer: A. ax + by ≥ c
Why: Constraints are linear inequalities, of the type ax + by ≤ c or ax + by ≥ c.
Q36.
If the objective is Z = x + y and a corner point is (3, 4), the value of Z there is:
Answer: A. 7
Why: Z = 3 + 4 = 7.
Q37.
For a minimisation LPP, we choose the corner point giving the ___ value of Z:
Answer: A. smallest
Why: Minimisation seeks the corner point with the least objective value.
Q38.
The feasible region of a linear programming problem is always a ___ set:
Answer: A. convex
Why: The intersection of half-planes is convex, so the feasible region is a convex set.
Q39.
If two corner points give the same optimal value of Z, then the optimum occurs at:
Answer: A. every point on the joining edge
Why: When two vertices tie, every point on the segment joining them is also optimal.
Q40.
The boundary lines of the constraints are obtained by replacing each inequality with:
Answer: A. an equality
Why: Turning ≤ or ≥ into = gives the boundary lines that are plotted.
Q41.
Minimize Z = 3x + 9y subject to x + 3y <= 60, x + y >= 10, x <= y, x >= 0, y >= 0. The corner points of the feasible region are (0,10), (5,5), and (15,15). Which gives the minimum Z?
Answer: B. (5,5) with Z = 60
Why: Corner-point method: Z(0,10) = 0+90 = 90; Z(5,5) = 15+45 = 60; Z(15,15) = 45+135 = 180. Smallest value is 60 at (5,5). Answer: minimum Z = 60 at (5,5).
Q42.
A furniture maker produces chairs (x) and tables (y). Each chair needs 2 hours of carpentry and 1 hour of finishing; each table needs 1 hour of carpentry and 3 hours of finishing. Only 40 carpentry hours and 60 finishing hours are available. If profit is Rs 30 per chair and Rs 60 per table, which is the correct constraint pair?
Answer: A. 2x + y <= 40 and x + 3y <= 60
Why: Carpentry: 2 hours per chair plus 1 hour per table, limited to 40 hours: 2x + y <= 40. Finishing: 1 hour per chair plus 3 hours per table, limited to 60 hours: x + 3y <= 60.
Q43.
For the furniture problem with constraints 2x + y <= 40, x + 3y <= 60, x >= 0, y >= 0 and objective Z = 30x + 60y, the corner points are (0,0), (20,0), (0,20), and the intersection of the two lines. Find the intersection point.
Answer: A. (12, 16)
Why: Solve 2x+y=40 and x+3y=60 together. From the first, y=40-2x. Substitute: x+3(40-2x)=60, x+120-6x=60, -5x=-60, x=12, y=40-24=16. Intersection: (12,16).
Q44.
Continuing the furniture problem (Z = 30x + 60y, corners (0,0), (20,0), (12,16), (0,20)), what is the maximum profit?
Answer: A. 1320
Why: Corner-point method: Z(0,0)=0; Z(20,0)=600; Z(12,16)=30(12)+60(16)=360+960=1320; Z(0,20)=0+1200=1200. Largest value is 1320 at (12,16). Answer: maximum profit = Rs 1320.
Q45.
A diet problem requires at least 8 units of vitamin A and at least 11 units of vitamin B daily. Food 1 (x units) gives 2 units of A and 1 unit of B per item; Food 2 (y units) gives 1 unit of A and 2 units of B per item. Which constraints model the minimum requirements?
Answer: A. 2x + y >= 8 and x + 2y >= 11
Why: Vitamin A requirement: 2 units from Food 1 plus 1 unit from Food 2 per item, total at least 8: 2x + y >= 8. Vitamin B: 1 unit from Food 1 plus 2 units from Food 2, total at least 11: x + 2y >= 11.
Q46.
For the diet problem with constraints 2x + y >= 8, x + 2y >= 11, x >= 0, y >= 0 and cost Z = 50x + 70y, the relevant corner points are (0,8), (5/3, 14/3), and (11,0). What is the minimum cost?
Answer: A. 410 at (5/3, 14/3)
Why: Z(0,8) = 50(0)+70(8) = 560. Z(11,0) = 50(11)+70(0) = 550. Z(5/3,14/3) = 50(5/3)+70(14/3) = 250/3+980/3 = 1230/3 = 410. The minimum cost is 410 at (5/3, 14/3).
Q47.
In an LPP to minimize cost with an unbounded feasible region (extending infinitely upward and to the right), which statement is generally true?
Answer: B. A minimum can exist at a corner point, since the region is bounded below-left even though unbounded elsewhere
Why: Even though the feasible region is unbounded in some directions, a minimum can still exist at a corner point if the region is closed off in the direction that would decrease the objective function further.
Q48.
Two corner points (10, 0) and (0, 15) of a feasible region both give Z = 300 for the objective function Z = 30x + 20y. What does this indicate about the LPP?
Answer: B. The LPP has multiple optimal solutions along the segment joining the two points
Why: Z(10,0)=300, Z(0,15)=300, both equal. This happens when the objective function's line is parallel to the boundary line joining these corners, giving infinitely many optimal solutions along that edge.
Q49.
A factory produces two products A and B. Each unit of A needs 3 machine-hours and 1 labour-hour; each unit of B needs 1 machine-hour and 2 labour-hours. Only 12 machine-hours and 10 labour-hours are available daily. Which constraint pair models this?
Answer: D. 3x + y <= 12 and x + 2y <= 10, x,y >= 0
Why: Machine-hours: 3 per unit of A plus 1 per unit of B, limited to 12, giving 3x + y <= 12. Labour-hours: 1 per A plus 2 per B, limited to 10, giving x + 2y <= 10.
Q50.
For the LPP Maximize Z = 5x + 4y subject to 3x + y <= 12 and x + 2y <= 10, x,y >= 0, the corner points include (0,0), (4,0), (0,5), and the intersection of the two boundary lines. Find Z at that intersection point.
Answer: C. 28.4
Why: Solving 3x+y=12 and x+2y=10 together gives x=2.8, y=3.6. Then Z = 5(2.8)+4(3.6) = 14+14.4 = 28.4, the maximum value of Z over the feasible region.
Q51.
Maximise Z = 5x + 3y subject to x + y ≤ 4, x ≥ 0, y ≥ 0 (corners (0,0), (4,0), (0,4)). The maximum Z is:
Answer: A. 20
Why: Z(4,0) = 20, Z(0,4) = 12, Z(0,0) = 0; the maximum is 20.
Q52.
For an unbounded feasible region, the maximum value of the objective function may:
Answer: A. fail to exist
Why: On an unbounded region the objective can increase without limit, so a maximum may not exist.
Q53.
In the LPP maximise Z = x + y subject to x + y ≤ 6, the maximum value of Z is:
Answer: A. 6
Why: Z = x + y is bounded above by 6, achieved anywhere on x + y = 6.
Q54.
The set of all feasible solutions of a two-variable LPP forms a:
Answer: A. convex polygon
Why: The feasible region bounded by linear constraints is a convex polygon.
Q55.
Minimise Z = 2x + 3y over the corner points (0,0), (2,0) and (0,2). The minimum value is:
Answer: A. 0
Why: Z(0,0) = 0, which is the smallest, so the minimum is 0.
Q56.
For the constraint 2x + y ≤ 10, the boundary line meets the y-axis (x = 0) at:
Answer: A. (0, 10)
Why: Setting x = 0 gives y = 10, so the line meets the y-axis at (0, 10).
Q57.
The point (2, 3) satisfies the constraint x + y ≤ 6 because:
Answer: A. 2 + 3 = 5, which is ≤ 6
Why: Since 2 + 3 = 5 and 5 ≤ 6, the point lies in the feasible region.
Q58.
The number of decision variables in a standard two-variable LPP is:
Answer: A. 2
Why: A two-variable LPP has decision variables x and y - two in all.
Q59.
In a typical diet problem, the objective function usually represents the ___ to be minimised:
Answer: A. cost
Why: Diet problems generally minimise the total cost of the food subject to nutritional constraints.
Q60.
Maximise Z = 4x + 3y over corners (0,0), (6,0), (0,4) and (3,3). The maximum value is:
Answer: A. 24
Why: Z(6,0) = 24, Z(3,3) = 21, Z(0,4) = 12; the maximum is 24.
Q61.
For an LPP with corner points (0,0), (4,0), (0,5) and (2,3), the maximum of Z = 3x + 4y is:
Answer: B. 20
Why: Evaluating: 0, 12, 20, 18. The maximum is 20 at (0, 5).
Q62.
The feasible region of any linear programming problem is always a:
Answer: A. convex set
Why: It is the intersection of half-planes, which is always convex.
Q63.
For corner points (0,0), (3,0) and (0,4), the maximum of Z = 2x + 3y is:
Answer: C. 12
Why: Values are 0, 6 and 12; the maximum is 12 at (0, 4).
Q64.
For corner points (2,3), (4,1) and (1,5), the minimum of Z = x + y is:
Answer: B. 5
Why: Values are 5, 5 and 6; the minimum is 5 (attained at two corners).
Q65.
Subject to 2x + y ≤ 10 with x, y ≥ 0, the maximum value of x is:
Answer: A. 5
Why: Setting y = 0 gives 2x ≤ 10, so the largest x is 5.
Q66.
If the optimal value of an LPP occurs at two adjacent corner points, then the problem has:
Answer: C. infinitely many optimal solutions
Why: The entire edge joining those corners is optimal, giving infinitely many solutions.
Q67.
For corner points (0,0), (0,8), (4,10), (6,8) and (6,0), the maximum of Z = 4x + 3y is:
Answer: B. 48
Why: Values: 0, 24, 46, 48, 24. The maximum is 48 at (6, 8).
Q68.
Subject to x + y ≤ 4 with x, y ≥ 0, the maximum of Z = x + 2y is:
Answer: C. 8
Why: The maximum is at (0, 4), giving Z = 8.