Linear Programming | ISC Class 12 Maths Notes

This note covers linear programming terminology, mathematical formulation, constraints, feasible regions, the graphical method, corner point results, maximum and minimum values, multiple optimal solutions, unbounded and infeasible problems, application areas, advantages and limitations.

What is a linear programming problem?

Optimisation means finding a maximum or minimum value, such as maximum profit, minimum cost or minimum use of resources. A linear programming problem, abbreviated to LPP, is a special optimisation problem involving a linear objective and linear restrictions on non-negative variables.

Definition: Linear programming finds the greatest or least value of a linear function while satisfying specified linear constraints and non-negative restrictions on the variables.

What do the symbols represent?

Let x and y denote the two quantities to be chosen. They are decision variables. Let Z denote the quantity to be optimised, and let a and b denote fixed numerical coefficients, meaning the numbers multiplying the variables.

The expression Z = ax + by is a linear objective function. Here ax means a multiplied by x, and by means b multiplied by y. The equals sign states equality. The objective specifies what is being maximised or minimised.

Constraints are restrictions expressed through linear inequalities or equations. The symbols ≤ and ≥ mean “less than or equal to” and “greater than or equal to”. The restrictions x ≥ 0, y ≥ 0 mean that neither decision variable can be negative.

The word linear describes the mathematical relationships. The word programming refers to determining a plan of action. Choosing permissible quantities and selecting the best permissible plan are different tasks: the constraints perform the first role, while the objective function supplies the criterion for the second.

How is a word problem translated into a mathematical model?

Begin by identifying the quantities that can be chosen and the restrictions on those choices. A mathematical model expresses the problem using variables, an objective function and constraints. Keep the units attached to the data while forming each expression.

How does the furniture problem become an LPP?

A furniture dealer has ₹50,000 to invest and storage for at most 60 pieces. Each table costs ₹2,500 and each chair ₹500. He estimates a profit of ₹250 per table and ₹75 per chair, assuming that he can sell all the items he buys.

For this problem, define x as the number of tables, y as the number of chairs and Z as total profit in rupees. The symbol ₹ denotes rupees. The decision variables now have specific meanings rather than merely representing unnamed quantities.

Dividing the investment inequality by 500 gives 5x + y ≤ 100. This simpler expression represents the same investment restriction. The storage inequality remains unchanged, since its terms count pieces rather than purchase cost.

A non-trivial constraint here means a restriction beyond the non-negative conditions. Investment and storage supply two such constraints. Do not turn “at most” into an equality: a permissible plan can leave some money or storage unused.

The model is complete only when it includes the instruction to maximise, the profit expression, both resource restrictions and both non-negative conditions. The assumption about selling all purchases is part of the problem and must accompany its interpretation.

How do feasible regions and feasible solutions differ?

The feasible region is the common set of points satisfying every constraint, including non-negativity. A feasible solution is an individual point in that set. An infeasible solution violates at least one constraint, and lies outside the feasible region.

A point is written as an ordered pair (x, y), with the first coordinate giving x and the second giving y. Points inside the feasible region and on its included boundary represent feasible solutions. Feasibility does not itself show that the objective is optimal.

How is the common region drawn?

A boundary line is obtained by replacing an inequality with equality. A half plane is one of the two sides into which a straight line divides the coordinate plane. For an inequality containing equality, its boundary is included.

Draw each boundary, select the side satisfying its inequality, and retain the common part. The restrictions x ≥ 0 and y ≥ 0 confine attention to the first quadrant, where both coordinates are non-negative, including the positive coordinate axes.

Here O, A, B and C name points, and O is the origin, the intersection of the coordinate axes. A vertex, or corner point, is a point of the feasible region where two boundary lines meet.

The shaded region is bounded by the coordinate axes and the lines 5x + y = 100 and x + y = 60. Its labelled vertices are O(0, 0), A(20, 0), B(10, 50) and C(0, 60).

See Fig. 12.1 in your NCERT textbook

In the furniture model, (10, 50), (0, 60) and (20, 0) are feasible. By contrast, (25, 40) is infeasible. Checking a proposed purchase against just the storage restriction is insufficient: the investment restriction and non-negativity must also hold.

Which results justify checking only the corner points?

Let R denote the feasible region. A region is bounded if it can be enclosed within a circle. Otherwise it is unbounded, extending indefinitely in some direction. An optimal feasible solution is a feasible point giving the greatest or least objective value sought.

Theorem: An existing optimum occurs at a corner

For the polygonal feasible regions considered here, if the linear objective Z = ax + by has an optimal value, that value occurs at a corner point of R. The condition if an optimal value exists is essential.

Theorem: A bounded feasible region has both extrema

If R is a non-empty bounded feasible region, Z = ax + by has both a maximum and a minimum on R, each attained at a corner point. Extrema means maximum and minimum values. “Non-empty” means that at least one feasible point exists.

Result: Equal optimal corner values extend along their joining segment

If two corner points give the same maximum value, every point on the line segment joining them gives that maximum. The corresponding statement holds for a shared minimum. A line segment contains the points between two endpoints, including those endpoints.

The feasible region is convex: the segment joining any two of its points remains inside it. This explains why the segment joining optimal corners remains feasible. Equal values at two corners must be identified as optimal before applying the result about multiple optima.

Note: If the feasible region is unbounded, a maximum or a minimum may not exist. However, if it exists, it occurs at a corner point in the problems considered here. Unboundedness does not by itself settle whether an optimum exists.

How does the corner point method solve the furniture problem?

The corner point method reduces the search over a feasible region to evaluating its vertices, followed by any check required for unboundedness. It is important to identify the region before evaluating the objective: an intersection outside the region is not a feasible corner.

What is the order of the method?

1. Plot the constraints and identify their common feasible region.
2. Find its vertices by inspection or by solving the equations of intersecting boundary lines.
3. Calculate the objective value at every vertex and compare the results.
4. For a bounded region, select the required greatest or least value. For an unbounded region, perform the additional half-plane check.

Worked example 1. Maximise the furniture dealer’s profit Z = 250x + 75y, where x counts tables and y counts chairs, subject to 5x + y ≤ 100, x + y ≤ 60, x ≥ 0 and y ≥ 0.

Answer: The bounded region has vertices (0, 0), (20, 0), (10, 50) and (0, 60). Subtracting x + y = 60 from 5x + y = 100 gives 4x = 40, hence x = 10 and y = 50.

Comparing the corner values gives a maximum profit of ₹6,250 at (10, 50). The dealer should buy 10 tables and 50 chairs, assuming he can sell all the items bought.

The table shows why buying tables alone is not the best plan, despite their greater profit per item. Buying chairs alone is also inferior. The combined resource constraints determine which mixtures can be considered, and the objective then compares them.

How is a maximum found on a bounded region?

For a non-empty bounded feasible region, the largest corner value is the required maximum. The objective function does not need to be largest where both decision variables are positive. A feasible point on a coordinate axis can be optimal.

How do intersecting constraints determine the corners?

Worked example 2. For non-negative decision variables x and y, maximise the objective Z = 4x + y subject to x + y ≤ 50 and 3x + y ≤ 90.

Answer: The boundary equations meet at (20, 30), obtained by subtracting them to give 2x = 40. The bounded feasible region has corners (0, 0), (30, 0), (20, 30) and (0, 50).

The corresponding objective values are 0, 120, 110 and 50. Therefore, the maximum value of Z is 120 at (30, 0).

The shaded quadrilateral has vertices O(0, 0), A(30, 0), B(20, 30) and C(0, 50). The plotted oblique boundaries are x + y = 50 and 3x + y = 90.

See Fig. 12.2 in your NCERT textbook

What happens when there is one resource constraint?

Worked example 3. Maximise the objective Z = 3x + 4y for decision variables x and y satisfying x + y ≤ 4, x ≥ 0 and y ≥ 0.

Answer: The feasible region is the triangle with vertices (0, 0), (4, 0) and (0, 4). Substitution gives Z = 0, 12 and 16 respectively. The maximum is 16 at (0, 4).

In both examples, use all the vertices of the common region. Do not assume the intersection of the two oblique lines must be optimal. The comparison of objective values, rather than the visual prominence of an intersection, decides the answer.

How is a minimum found on a bounded region?

Minimisation uses the same feasible-region construction as maximisation. The change is in the final comparison: select the smallest objective value. A lower-bound constraint can exclude the origin, even when both decision variables are non-negative.

How can upper and lower restrictions form a triangle?

Worked example 4. Minimise the objective Z = 200x + 500y for decision variables x and y subject to x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0 and y ≥ 0.

Answer: The bounded feasible triangle has vertices (0, 5), (4, 3) and (0, 6). To find the intersection, double x + 2y = 10, then subtract it from 3x + 4y = 24. This gives x = 4 and y = 3.

The values of Z at these vertices are 2500, 2300 and 3000. Therefore, the minimum value is 2300 at (4, 3).

The shaded triangle has vertices A(0, 5), B(4, 3) and C(0, 6). Its oblique boundaries are x + 2y = 10 and 3x + 4y = 24; its remaining side lies on the vertical axis.

See Fig. 12.3 in your NCERT textbook

The origin is infeasible because it fails x + 2y ≥ 10. Its objective value therefore has no place in the corner comparison. A low value found outside the feasible region cannot be used to answer a constrained minimisation problem.

This example also illustrates why the direction of each inequality matters. One restriction sets a lower requirement and the other an upper limit. The feasible triangle is their common part after imposing non-negativity, not the entire region below either line separately.

When does a problem have multiple optimal solutions?

Multiple optimal solutions are different feasible points giving the same optimal objective value. A corner table can reveal this situation when two vertices share the greatest or least value. The joining segment must then be included in the description of the solutions.

How can both extrema be found in one problem?

Worked example 5. For decision variables x and y, minimise and maximise the objective Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0 and y ≥ 0.

Answer: The feasible region is bounded, with vertices A(0, 10), B(5, 5), C(15, 15) and D(0, 20). Here A, B, C and D label its successive corners.

The corner values are 90, 60, 180 and 180. The minimum is 60 at B. The maximum is 180 at C and D and at every point on the segment joining them.

The shaded quadrilateral ABCD lies between the labelled boundaries x + y = 10, x = y, x + 3y = 60 and the vertical axis. The maximum occurs along the side joining C(15, 15) to D(0, 20).

See Fig. 12.4 in your NCERT textbook

Along that side, x + 3y = 60. Since Z = 3(x + 3y), every point on it gives Z = 180. The algebra confirms why listing only the two endpoints would omit other optimal solutions.

This problem contains three non-trivial constraints in addition to non-negativity. Each contributes to the region. In particular, x ≤ y selects the side of the line x = y on which the first coordinate does not exceed the second.

How must an unbounded feasible region be checked?

An unbounded region can have feasible points beyond every finite drawing. Consequently, the greatest or least value in a corner table is initially a candidate optimum, meaning a value still requiring confirmation. A maximum or minimum may not exist.

What is the additional half-plane test?

Let M be the largest corner value and m the smallest. The symbols > and < mean strictly greater than and strictly less than. An open half plane excludes its boundary line.

For maximisation, check whether ax + by > M has any point in common with the feasible region. If it has none, M is the maximum. If it has common points, the objective has no maximum in this corner point procedure.

For minimisation, check ax + by < m. If this open half plane has no feasible point, m is the minimum. If it has common points, the objective has no minimum. Do not replace either strict inequality with an equality.

Worked example 6. Using − for subtraction or a negative number, minimise the objective Z = −50x + 20y for decision variables x and y satisfying 2x − y ≥ −5, 3x + y ≥ 3, 2x − 3y ≤ 12, x ≥ 0 and y ≥ 0.

Answer: The feasible region is unbounded. Its corner values are shown below. Although −300 is the smallest of these, the open half plane −50x + 20y < −300 has points in common with the feasible region. Therefore, Z has no minimum.

The shading continues upwards and rightwards beyond the displayed corners. A dashed line labelled −5x + 2y = −30 passes through (6, 0), with shaded feasible points on the side where the objective is smaller.

See Fig. 12.5 in your NCERT textbook

Can an unbounded region still have a minimum?

Worked example 7. Minimise the objective Z = 3x + 5y for decision variables x and y subject to x + 3y ≥ 3, x + y ≥ 2, x ≥ 0 and y ≥ 0.

Answer: Using a slash to denote division, the unbounded region has corners (0, 2), (3/2, 1/2) and (3, 0), giving Z = 10, 7 and 9. Thus 3/2 means three divided by two.

Using × for multiplication, Z = 2(x + y) + (x + 3y) and the constraints give Z ≥ 2 × 2 + 3 = 7. Thus Z < 7 has no feasible point. The minimum is 7 at (3/2, 1/2).

How is an infeasible problem recognised?

A problem is infeasible when no point satisfies all its constraints simultaneously. There is no feasible region to search and no optimal feasible solution. This differs from an unbounded feasible problem, which does contain permissible points.

How can individually possible restrictions conflict?

Worked example 8. Minimise the objective Z = 3x + 2y for decision variables x and y subject to x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0 and y ≥ 0.

Answer: Since y ≥ 0, we have 3x + 5y ≥ 3(x + y) ≥ 24. This contradicts 3x + 5y ≤ 15. The selected half planes have no common point in the first quadrant, so there is no feasible solution or minimum.

The graph shows x + y = 8 through (0, 8) and (8, 0), and 3x + 5y = 15 through (0, 3) and (5, 0). Their selected shaded regions do not overlap.

See Fig. 12.6 in your NCERT textbook

The algebra and graph support the same conclusion. Each inequality separately describes a possible set of points, but their requirements conflict when imposed together. Feasibility concerns the common intersection of all selected regions, not the existence of points in each separate region.

There is no need to calculate an objective table once infeasibility is established. Evaluating the objective at line intersections would not create a feasible solution. The restrictions must be jointly consistent before the question of an optimum can arise.

Keep three conclusions distinct: a feasible region can be bounded; it can be unbounded; or it can be absent. “Unbounded” describes a region that exists and extends indefinitely. “No feasible region” describes the failure of the constraints to admit any common point.

What types of problems and application areas use linear programming?

Linear programming has applications in industry, commerce and management science. The same mathematical framework can describe different decisions: choose quantities, respect restrictions and optimise a stated linear objective. The interpretation changes with the problem, even when the solution method remains the same.

How do maximisation and minimisation differ?

Maximisation problems seek the greatest attainable objective value, as in the furniture dealer’s profit problem. Minimisation problems seek the least attainable value, such as cost or resource use. Neither instruction changes the need to satisfy all constraints.

The question must tell us what quantity is being optimised. Maximising profit is not interchangeable with minimising purchase cost. In the furniture example, purchase costs appear in the investment constraint, whereas profits appear in the objective function.

What are common application types?

• Manufacturing or production: decide quantities of products while respecting restrictions on available resources, commonly with profit as the objective.
• Diet: choose quantities of foods to satisfy nutritional requirements, commonly while minimising total cost.
• Transportation: choose quantities to send from supply locations to destinations while meeting supply and demand restrictions, commonly while minimising transport cost.

These descriptions identify application types, not ready-made equations. The actual objective coefficients and constraint coefficients must come from the problem’s data. The choice of variables must also specify what each quantity counts or measures.

Before drawing a graph, translate every relevant restriction. A problem can resemble a familiar application while requiring different inequality directions. A maximum available amount and a minimum required amount impose different conditions, so the wording controls the formulation.

What are the advantages and limitations of linear programming?

Linear programming makes the relationship between a decision, its restrictions and its objective explicit. Its usefulness depends on how well the mathematical model represents the actual situation. A solution is optimal for the stated model and its assumptions.

What advantages does the method offer?

• Systematic comparison: the objective supplies one stated criterion for comparing permissible plans, instead of choosing from isolated trial cases.
• Resource allocation: constraints represent limited resources, allowing alternative uses of those resources to be considered together.
• Clear restrictions: the feasible region separates admissible choices from choices that violate one or more conditions.
• Efficient graphical search: the corner point results reduce the search over a polygonal region to its vertices, with an additional check when the region is unbounded.

What limitations must be remembered?

• Linearity: the objective and constraints must be represented by linear relationships. A relationship that is not linear cannot be inserted unchanged into this model.
• Dependence on data: the answer uses the specified numerical coefficients and assumptions. It does not establish that estimated profits will actually be realised.
• Indivisible quantities: the graphical model allows real-valued coordinates. If a quantity must be a whole number, a fractional solution needs separate attention to that restriction.
• Graphical scope: the plane-graph method used here handles two decision variables. Problems with more variables require other methods.

The furniture problem illustrates the role of assumptions particularly clearly. Its estimated profits support the chosen objective, and its conclusion assumes that all purchased furniture can be sold. Keeping that qualification preserves the meaning of the optimum.

Finally, finding no feasible solution or no finite optimum is a mathematical conclusion, not a reason to choose an arbitrary corner. The stated constraints and objective determine what conclusion the method supports.

Glossary

• Linear programming — A method for optimising a linear objective under linear constraints and non-negative variable restrictions.
• Decision variables — Quantities whose values are chosen to determine a feasible plan of action.
• Objective function — The linear expression whose maximum or minimum value the problem seeks.
• Constraints — Equations, inequalities or restrictions that the decision variables must satisfy together.
• Non-negative restrictions — Conditions requiring decision variables to be zero or positive rather than negative.
• Feasible region — The common set of points satisfying every constraint, including non-negative restrictions.
• Feasible solution — A point inside or on the included boundary of the feasible region.
• Infeasible solution — A point violating at least one constraint and lying outside the feasible region.
• Optimal feasible solution — A feasible point giving the required maximum or minimum objective value.
• Corner point — A point in the feasible region at which two boundary lines intersect.
• Bounded region — A feasible region that can be enclosed completely within a circle.
• Unbounded region — A feasible region extending indefinitely in some direction, which cannot be enclosed within a circle.
• Convex region — A region containing the entire line segment joining any two of its points.
• Multiple optimal solutions — Different feasible points at which the same optimal objective value is attained.

Common errors and misconceptions

• Misconception: Any point satisfying one constraint is feasible. Correct: It must satisfy all constraints, including non-negativity.
• Misconception: The objective function is another resource constraint. Correct: It is the quantity to be maximised or minimised over the feasible region.
• Misconception: “At most” requires equality. Correct: It allows values below the stated limit as well as equality.
• Misconception: Every pairwise line intersection is a feasible corner. Correct: The point must also lie in the common feasible region.
• Misconception: An unbounded region cannot have any optimum. Correct: An optimum may or may not exist; apply the additional check.
• Misconception: The smallest corner value is sufficient in every minimisation problem. Correct: For an unbounded region, test whether a smaller-value open half plane contains feasible points.
• Misconception: Two optimal corners mean exactly two optimal solutions. Correct: Every point on their joining segment also gives the same optimal value.
• Misconception: Infeasible and unbounded mean the same thing. Correct: Infeasibility means no common feasible point; an unbounded feasible region contains points and extends indefinitely.

Exam-style questions with model answers

Q1. Distinguish a feasible solution from an optimal feasible solution in a linear programming problem. [2 marks]

1. A feasible solution is a point satisfying every constraint, including the non-negative restrictions on the decision variables.
2. An optimal feasible solution is a feasible point that gives the required maximum or minimum value of the objective function.

Q2. A dealer can invest ₹50,000 and store at most 60 furniture items. A table costs ₹2,500 and a chair ₹500. Estimated profits are ₹250 per table and ₹75 per chair. Assuming all purchases can be sold, formulate the profit-maximisation problem. [4 marks]

1. Let x be the number of tables and y the number of chairs purchased. Let Z represent the total estimated profit in rupees.
2. The objective is to maximise Z = 250x + 75y, using the stated profit earned on each type of item.
3. The investment restriction is 2500x + 500y ≤ 50000, equivalently 5x + y ≤ 100.
4. The storage restriction is x + y ≤ 60. Include x ≥ 0 and y ≥ 0, since the purchase quantities cannot be negative.

Q3. Maximise the objective Z = 4x + y for decision variables x and y subject to x + y ≤ 50, 3x + y ≤ 90, x ≥ 0 and y ≥ 0. Identify the graphical feasible region and show the corner point calculation. [5 marks]

1. Draw the boundary lines x + y = 50 and 3x + y = 90. The non-negative restrictions confine the feasible region to the first quadrant, including its axes.
2. Keep the common region satisfying both upper-bound inequalities. It is a bounded quadrilateral, so its maximum can be found by comparing the corner values.
3. Subtracting the boundary equations gives 2x = 40. Therefore, their intersection is (20, 30), and the full corner list is (0, 0), (30, 0), (20, 30), (0, 50).
4. Substitute each corner into Z = 4x + y. The corresponding objective values are 0, 120, 110 and 50 respectively.
5. The greatest of these values is 120. Hence the required maximum is Z = 120, attained when x = 30 and y = 0.

Q4. Minimise the objective Z = 200x + 500y for decision variables x and y subject to x + 2y ≥ 10, 3x + 4y ≤ 24, x ≥ 0 and y ≥ 0. Show the graphical region and corner comparison. [5 marks]

1. Draw x + 2y = 10 and 3x + 4y = 24. Select the sides satisfying the given inequalities and retain their common part in the first quadrant.
2. On the vertical axis the feasible boundary runs from (0, 5) to (0, 6). The two oblique boundaries meet at (4, 3), found by solving their equations together.
3. These three points form the bounded feasible triangle. The bounded-region theorem therefore permits the minimum to be determined by evaluating the objective at its corners.

   The shaded triangle has vertices A(0, 5), B(4, 3) and C(0, 6). Its oblique boundaries are x + 2y = 10 and 3x + 4y = 24; its remaining side lies on the vertical axis.

   See Fig. 12.3 in your NCERT textbook

4. At (0, 5), (4, 3) and (0, 6), the objective values are 2500, 2300 and 3000 respectively.
5. The smallest value is 2300. Therefore, Z has minimum value 2300 at the feasible solution x = 4, y = 3.

Q5. For decision variables x and y, the objective is Z = 3x + 9y. The constraints are x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0 and y ≥ 0. The bounded feasible region has corners (0, 10), (5, 5), (15, 15) and (0, 20). Find both extrema and describe every maximum solution. [4 marks]

1. The objective values at the stated corners are 90, 60, 180 and 180 respectively, obtained by substitution into Z = 3x + 9y.
2. The minimum value is 60, attained at the feasible corner (5, 5).
3. The maximum value is 180, attained at both (15, 15) and (0, 20).
4. Every point on the segment joining these maximum corners also attains 180. Along this segment x + 3y = 60, so Z = 3(x + 3y) = 180.

Q6. Minimise the objective Z = 3x + 5y for decision variables x and y under x + 3y ≥ 3, x + y ≥ 2, x ≥ 0 and y ≥ 0. The unbounded feasible region has corners (0, 2), (3/2, 1/2) and (3, 0), where a slash denotes division. Verify whether its smallest corner value is a true minimum. [3 marks]

1. Substitution gives objective values 10, 7 and 9 at the three supplied corners respectively. Thus 7 is the candidate minimum.
2. Rewrite the objective as Z = 2(x + y) + (x + 3y). The constraints imply Z ≥ 2 × 2 + 3 = 7 at every feasible point.
3. Therefore, no feasible point lies in the open half plane Z < 7. The minimum is 7, attained at (3/2, 1/2), despite the region being unbounded.

Q7. For decision variables x and y, consider minimising Z = 3x + 2y subject to x + y ≥ 8, 3x + 5y ≤ 15, x ≥ 0 and y ≥ 0. Explain why there is no feasible solution. [3 marks]

1. Since y is non-negative, 3x + 5y = 3(x + y) + 2y is at least 3(x + y).
2. The constraint x + y ≥ 8 consequently requires 3x + 5y ≥ 24. This contradicts the separate restriction 3x + 5y ≤ 15.
3. No point can satisfy all these restrictions together. Hence the feasible region is empty, and the stated minimisation problem has no optimal feasible solution.

Key takeaways

• A complete linear programming formulation specifies decision variables, the objective to optimise, every linear constraint and the non-negative restrictions.
• The feasible region is the common set satisfying all constraints; each point in it represents a feasible solution.
• For a non-empty bounded feasible region, both maximum and minimum objective values occur at corner points.
• On an unbounded region, a maximum or minimum may not exist, so corner values require an additional half-plane check.
• When two corners share an optimal value, every point on their joining segment also gives that same value.
• An infeasible problem has no common feasible point, while an unbounded feasible problem has permissible points extending indefinitely.
• Interpret an optimal solution using the original quantities, units and assumptions, including any assumption about selling all purchased items.
• Advantages include systematic comparison and resource allocation; limitations include linearity, dependence on data and attention to indivisible quantities.

Test yourself

What does “programming” mean in linear programming?

It means determining a programme or plan of action satisfying the stated restrictions.

Does a point on the included boundary of a feasible region count as feasible?

Yes. Points inside the region and on its included boundary satisfy the constraints.

What distinguishes a bounded feasible region from an unbounded one?

A bounded region can be enclosed within a circle; an unbounded region extends indefinitely in some direction.

Why is a table of corner values insufficient for an unbounded region?

Feasible points beyond the displayed corners may yield better values, so the appropriate strict half-plane inequality must also be checked.

What follows when two feasible corners give the same maximum objective value?

Every point on the line segment joining those corners also gives the same maximum value.

Can an objective value at an infeasible point answer the optimisation problem?

No. The required optimal solution must satisfy every constraint as well as optimise the objective.

In an unbounded minimisation problem, what does an empty intersection with the strict lower-value half plane establish?

It establishes that no feasible point improves on the candidate, so the candidate is the minimum.

What should accompany the numerical answer to a word problem?

State what each decision variable represents, give the objective value with its units, and preserve the problem’s assumptions.