NCERT Plus Two (Class 12) Mathematics • Chapter 12

Linear Programming

Q: Define 'Feasible Region' and 'Optimal Solution' in a Linear Programming Problem.

Feasible Region: The common region determined by all the linear constraints including the non-negativity constraints ( x ≥ 0, y ≥ 0 ) of an LPP. Optimal Solution: Any point within the feasible region that gives the maximum or minimum value of the objective function Z = ax + by .

Q: State the Corner-Point Theorem for a bounded feasible region, and explain when an LPP admits infinitely many optimal solutions.

If the feasible region R is bounded, the objective function Z = ax + by attains both a maximum and a minimum value at corner points (vertices) of R . If two adjacent corner points produce the exact same optimal value of Z , then every point on the line segment joining them gives that same optimal value — resulting in infinitely many optimal solutions.

Q: A merchant sells two models of fans A and B. He has space for at most 60 fans and ₹50,000 to invest. Fan A costs ₹2,500 (profit ₹250 ) and Fan B costs ₹500 (profit ₹75 ). Formulate this as an LPP to maximise profit.

Let x = number of Fan A and y = number of Fan B. Maximise: Z = 250x + 75y Subject to: x + y ≤ 60 (storage), 2500x + 500y ≤ 50000 ⇒ 5x + y ≤ 100 (budget), and x ≥ 0, y ≥ 0 .

Q: Solve graphically: Maximise Z = 3x + 2y subject to x + 2y ≤ 10 , 3x + y ≤ 15 , x ≥ 0, y ≥ 0 .

Boundary lines: x + 2y = 10 through (10,0), (0,5) ; 3x + y = 15 through (5,0), (0,15) . Both shade toward (0,0) . Intersection: 3(10 − 2y) + y = 15 ⇒ 30 − 5y = 15 ⇒ y = 3, x = 4 . Corner points of bounded region: (0,0) → 0 , (5,0) → 15 , (4,3) → 12 + 6 = 18 , (0,5) → 10 . Hence Maximum Z = 18 at (4, 3) .

Q: Minimise Z = −3x + 4y subject to x + 2y ≤ 8 , 3x + 2y ≤ 12 , x ≥ 0, y ≥ 0 .

Lines intersect at: subtracting x + 2y = 8 from 3x + 2y = 12 gives 2x = 4 ⇒ x = 2, y = 3 . Bounded vertices: (0,0) → 0 , (4,0) → −12 , (2,3) → −6 + 12 = 6 , (0,4) → 16 . Minimum value of Z is −12 at (4, 0) .

Q: Solve graphically: Minimise Z = 3x + 5y such that x + 3y ≥ 3 , x + y ≥ 2 , x ≥ 0, y ≥ 0 .

Line x + 3y = 3 meets axes at (3,0), (0,1) ; Line x + y = 2 meets axes at (2,0), (0,2) . Both shade AWAY from (0,0) → Unbounded feasible region . Intersection: (x + 3y) − (x + y) = 3 − 2 ⇒ 2y = 1 ⇒ y = ½, x = 3/2 . Corner points: (3, 0) → 9 , (3/2, 1/2) → 9/2 + 5/2 = 7 , (0, 2) → 10 . Open Half-Plane Test: Graph dashed 3x + 5y < 7 (passes through (7/3, 0) and (0, 7/5) , shades toward (0,0) ). It has NO point in common with the feasible region. Therefore, Minimum Z = 7 at (3/2, 1/2) .

Q: Minimise and Maximise Z = 3x + 9y subject to x + 3y ≤ 60 , x + y ≥ 10 , x ≤ y , x ≥ 0, y ≥ 0 .

Bounded feasible region has 4 corner points: • A(0, 10) (on x=0, x+y=10 ) → Z = 90 • B(5, 5) (on x=y, x+y=10 ) → Z = 15 + 45 = 60 (Minimum!) • C(15, 15) (on x=y, x+3y=60 ) → Z = 45 + 135 = 180 (Maximum!) • D(0, 20) (on x=0, x+3y=60 ) → Z = 180 (Maximum!) Conclusion: Minimum Z = 60 at (5, 5) ; Maximum Z = 180 at every point on the line segment joining (15, 15) and (0, 20) .

Q: Determine graphically whether Z = −50x + 20y has a minimum value subject to 2x − y ≥ −5 , 3x + y ≥ 3 , 2x − 3y ≤ 12 , x ≥ 0, y ≥ 0 .

The shaded feasible region in Quadrant I is unbounded with corner points (0, 5) , (0, 3) , (1, 0) , and (6, 0) . Corner values: Z(0,5) = 100 , Z(0,3) = 60 , Z(1,0) = −50 , Z(6,0) = −300 . Open Half-Plane Check: Graph dashed −50x + 20y < −300 (i.e. −5x + 2y < −30 ). Test feasible point (10, 4) : −500 + 80 = −420 < −300 is TRUE! Since the open half-plane has common points with the feasible region, Z has NO minimum value .

Objective function, linear constraints, graphical half-planes, feasible regions, corner-point theorem & special LPP cases.

📄 Official NCERT Textbook PDF

Topic 12.2: Objective Function Z = ax + by, Linear Constraints & Decision Variables

Imagine Ananya opening a small furniture workshop in Kochi making wooden tables (x) and chairs (y)! She only has ₹50,000 in cash and 60 square metres of storage space, yet she wants to earn the highest possible profit! Turning a real-world "best profit or lowest cost within limits" puzzle into clean straight-line algebra is called Linear Programming (LPP)!

DECISION VARIABLESQuantities you choose: x, y (always non-negative: x ≥ 0, y ≥ 0!)

OBJECTIVE FUNCTIONLinear goal Z = ax + by (a, b constants) to Maximise or Minimise!

FEASIBLE & OPTIMALFeasible Region: points satisfying ALL constraints.
Optimal Solution: feasible point giving best Z!

🧠 Memory Trick: Why is it called LINEAR Programming? Every variable x, y appears ONLY to power 1 — no x², no xy, no 1/x anywhere!

💡 Simple Kid-Check: Every Linear Programming Problem works just like packing a school picnic bag: Decision Variables = how many sandwiches (x) and juice packs (y) to pack; Constraints = weight & budget limits; Objective Function Z = total yummy happiness to maximise!

1
Decision Variables (x, y): Unknown non-negative quantities we control.
2
Objective Function (Z = ax + by): The linear target formula we want to Maximise (profit) or Minimise (cost).
3
Linear Constraints + Non-Negativity: System of inequalities (≤ or ≥) PLUS x ≥ 0, y ≥ 0 restricting everything to Quadrant I!

LPP Term	Plain Meaning	Furniture Shop Example
Decision Variables	What we decide to produce/buy	`x` tables, `y` chairs
Objective Function	Target score `Z = ax + by`	Maximise profit `Z = 250x + 75y`
Constraints	Limited money, time, or space	`5x + y ≤ 100`, `x + y ≤ 60`
Non-negativity	Cannot make negative items!	`x ≥ 0, y ≥ 0`

NEVER forget to write the non-negativity constraints x ≥ 0, y ≥ 0 at the bottom of your LPP formulation! Without them, negative tables or negative chairs become "legal", letting the shaded region spill out of Quadrant I and costing you 1 mark in Kerala DHSE exams!

Topic 12.2.1: Translating Real-Life Business & Diet Problems into Constraints

Word problems in LPP look long only until you learn the secret English-to-Math dictionary! Whenever a story says "maximum available", "budget", or "at most", write ≤! Whenever it says "minimum requirement" or "at least", write ≥! Put everything into a tidy 2-column table, and the inequalities practically write themselves!

AT MOST / BUDGET"Maximum", "not exceeding", "available", "at most" → use ≤!

AT LEAST / QUOTA"Minimum", "requires", "at least", "not less than" → use ≥!

TABLE-FIRST RULEPut decisions (x, y) as columns and resources/nutrients as rows — read across!

🧠 Memory Trick: Resource Ceiling (cash, machine hours, storage) = ≤; Demand Floor (vitamins, daily target) = ≥!

💡 Simple Kid-Check: Never try to write equations directly from a giant paragraph! Draw a quick 2-column box for Item x and Item y, fill in numbers row-by-row (Cost row, Time row, Vitamin row), and attach ≤ or ≥ to the right-hand limit!

1
Identify x, y: Ask "What two items am I choosing how many/how much to make or buy?"
2
Write Objective Z: Spot profit per unit (Maximise) or cost per unit (Minimise).
3
Build One Inequality Per Limit: Match units carefully (all minutes or all hours; all ₹ or all kg!).
4
Tack on x ≥ 0, y ≥ 0: Always finish with non-negativity!

🧒 Quick Diet Mix Translation:
Mix x kg of Food P (₹60/kg) & y kg of Food Q (₹80/kg) so cost is minimum. Food P has 3 units/kg Vitamin A & 5 units/kg Vitamin B; Food Q has 4 units/kg Vitamin A & 2 units/kg Vitamin B. Daily need: at least 8 units Vit A and 11 units Vit B.
• Minimise Cost: Z = 60x + 80y
• Vit A ("at least 8"): 3x + 4y ≥ 8 | Vit B ("at least 11"): 5x + 2y ≥ 11 | x ≥ 0, y ≥ 0!

Watch TWO classic word-problem traps! (1) Flipping ≤ and ≥: "Diet must contain at least 8 units" means ≥ 8 (you can have MORE than 8, not less!). (2) Unit mismatch: Machine A takes 30 minutes while availability is 4 hours — convert 4 hours = 240 minutes FIRST so both sides speak the same unit!

Topic 12.2.2: Graphical Plotting of Half-Planes & Feasible Region Shading

Every linear inequality like x + 2y ≥ 10 acts like a straight security fence slicing the graph sheet into two halves — one "allowed half-plane" and one "forbidden half-plane"! Draw every fence cleanly using just two intercepts, send the point (0, 0) as your scout to check which side passes, and the glowing overlap in Quadrant I is your Feasible Region!

2-INTERCEPT LINE PLOTReplace ≤/≥ by =. Set x=0 (get y-intercept) & y=0 (get x-intercept) — join with a SOLID line!

ORIGIN (0, 0) SCOUT TESTPlug (0, 0) into inequality:
TRUE → shade toward (0, 0) | FALSE → shade away from (0, 0)!

QUADRANT I LOCKx ≥ 0, y ≥ 0 forces shading strictly into the first quadrant (top-right)!

🧠 Memory Trick: "(0, 0) passes? Pull towards Origin! (0, 0) fails? Push away from Origin!" And if the line passes RIGHT THROUGH (0, 0), test (1, 0) or (0, 1) instead!

💡 Simple Kid-Check: Think of shading like putting up garden fences! Fence 1 (3x + 4y ≤ 24) allows the side containing (0, 0) because 0 ≤ 24 is TRUE. Fence 2 (x + 2y ≥ 10) blocks (0, 0) because 0 ≥ 10 is FALSE — so it pushes outward away from (0, 0)! Where BOTH fences agree in Quadrant I sits your Feasible Region!

1
Draw Boundary Lines: Find (a, 0) and (0, b) for each equation and draw a solid line (because ≤ and ≥ include the boundary!).
2
Test One Side: Plug (0, 0) into each inequality (or (1, 0) if the line hits (0, 0) like x − y ≤ 0).
3
Shade ONLY Common Intersection: Draw small arrows on each line first, then cleanly shade ONLY the common region satisfying ALL constraints!

🧒 Quick Plotting Check: Plot x + 2y ≥ 10 and 3x + 4y ≤ 24 (x, y ≥ 0).
• Line 1 (x + 2y = 10): meets axes at (10, 0) & (0, 5). Test (0,0): 0 ≥ 10 is FALSE → shade AWAY from (0,0).
• Line 2 (3x + 4y = 24): meets axes at (8, 0) & (0, 6). Test (0,0): 0 ≤ 24 is TRUE → shade TOWARD (0,0).
• Common region in Quadrant I is the crisp triangle with vertices (0, 5), (4, 3), (0, 6)!

Watch TWO sneaky graphical blunders! (1) Testing (0, 0) on a line passing through the origin like x − 2y ≥ 0 (or x ≤ y)! Plugging (0,0) gives 0 ≥ 0 — completely useless for picking a side! Pick (1, 0) or (5, 0) instead! (2) Never draw dashed lines for constraints — LPP constraints always use ≤ or ≥ (solid lines!).

Topic 12.2.2: Corner-Point Method & Fundamental Theorem of LPP

Once you shade a bounded feasible polygon, it holds infinitely many points inside! Surely you don't need to test millions of points to find the maximum of Z = 4x + y?! Relax! Slide a straight ruler representing Z = 4x + y across the polygon — the LAST place it touches before leaving is ALWAYS one of the sharp Corner Points (Vertices)!

FUNDAMENTAL THEOREMAny optimal value (max or min) of Z = ax + by MUST occur at a corner point (vertex) of the feasible region!

BOUNDED GUARANTEEIf the feasible region is bounded (closed polygon), BOTH Maximum & Minimum of Z are 100% guaranteed to exist!

4-STEP CORNER RECIPE1. Plot & shade region → 2. Find ALL vertices → 3. Make (x,y) | Z table → 4. Pick Max/Min!

🧠 Memory Trick: Optimal solutions NEVER hide in the mushy middle — they ALWAYS sit on a sharp corner vertex!

💡 Simple Kid-Check: Imagine tilting a tray full of marbles inside a fenced polygonal room — every marble rolls straight into one of the sharp corners! That's literally the Corner-Point Method: solve the boundary line intersections to get your 3–5 corner points, plug ONLY those into Z = ax + by, and crown the winner!

1
Find All Vertices: Read axis intercepts directly from your graph and solve simultaneous equations ONLY for interior intersections where two boundaries cross!
2
Build the Corner-Point Table: List every corner A, B, C, ... alongside its exact Z = ax + by value.
3
State the Conclusion Clearly: Write BOTH the optimal value of Z AND the exact point (x, y) where it occurs!

🧒 NCERT Example 1 (Classic 4-Mark Board Solution):
Maximise Z = 4x + y subject to x + y ≤ 50, 3x + y ≤ 90, x ≥ 0, y ≥ 0.
• Line 1 meets axes at (50, 0), (0, 50); Line 2 at (30, 0), (0, 90). Both shade toward (0,0).
• Intersection of x + y = 50 and 3x + y = 90: subtract → 2x = 40 ⇒ x = 20, y = 30.
• Corner-Point Table (Bounded Region):
(0, 0) → Z = 0 | (30, 0) → Z = 120 (MAX!) | (20, 30) → Z = 110 | (0, 50) → Z = 50.
• Answer: Maximum value of Z is 120 at the point (30, 0)!

NEVER include a fake intersection point that sits OUTSIDE the shaded feasible region! In Example 1 above, (0, 90) is the y-intercept of 3x + y = 90, yet it violates x + y ≤ 50 (90 > 50) — so it is NOT a corner of the feasible region! Always verify every vertex satisfies ALL constraints!

Topic 12.2.2: Bounded vs Unbounded Feasible Region (Open Half-Plane Test)

Diet and cost-minimisation problems often use ≥ constraints, creating an Unbounded feasible region that stretches endlessly outward! Suppose your corner-point table gives a smallest value m = 7 (or m = −300). Careful — since the region runs off to infinity, could some faraway point drop EVEN LOWER than m?! You MUST run one extra detective check: draw the dashed Open Half-Plane ax + by < m!

BOUNDED vs UNBOUNDEDBounded: fits inside a circle (optimal guaranteed).
Unbounded: stretches to ∞ (extra test REQUIRED!).

OPEN HALF-PLANE TESTFor Min with smallest corner m: graph ax + by < m (dashed!).
For Max with largest corner M: graph ax + by > M!

VERDICT RULENO point in common with feasible region → Optimal EXISTS!
ANY overlap → NO optimal solution!

🧠 Memory Trick: Draw ax + by < m as a DASHED challenger line! If the challenger touches the feasible region anywhere, your corner loses — NO minimum exists!

💡 Simple Kid-Check: Think of your best corner value m as the reigning champion! In a bounded room, no outsider exists. In an unbounded open field running to infinity, we send a challenger zone ax + by < m (strict inequality → dashed line!). Does ANY part of that challenger zone step onto the feasible field? If NO → champion m wins! If YES → someone out toward infinity beats m forever, so NO minimum exists!

1
Compute Corner Table First: Identify smallest corner value m (for Minimise) or largest M (for Maximise).
2
Draw Dashed Line ax + by = m: Test a sample point (e.g. (0, 0)) to shade the strict open half-plane ax + by < m (or > M).
3
Check Overlap with Feasible Region: Zero common points = m is the true minimum; even 1 overlapping point = NO minimum exists!

⚖️ Two Contrasting NCERT Classics Side-by-Side:
• Case A — Unbounded WITH Minimum (Ex 12.1 Q4): Minimise Z = 3x + 5y subject to x + 3y ≥ 3, x + y ≥ 2, x,y ≥ 0.
Corners: (3,0)→9, (3/2, 1/2)→7, (0,2)→10. Graph dashed 3x + 5y < 7: it stays strictly BELOW the region → NO common point ⇒ Minimum IS 7 at (3/2, 1/2)!
• Case B — Unbounded with NO Minimum (NCERT Ex 4): Minimise Z = −50x + 20y subject to 2x−y ≥ −5, 3x+y ≥ 3, 2x−3y ≤ 12, x,y ≥ 0.
Smallest corner is (6, 0) → −300. Graph dashed −50x + 20y < −300: test (10, 4) (feasible & Z = −420 < −300!) → Overlaps feasible region ⇒ NO minimum exists!

THE #1 reason Kerala Plus Two students lose 1–2 marks in a 4-mark LPP question! In an unbounded problem, writing "Minimum is 7" directly from the corner table WITHOUT drawing and testing the dashed open half-plane ax + by < m gets docked marks even when 7 happens to be right! And remember: strict inequalities < / > MUST be drawn as DASHED lines!

Topic 12.2.2: Special Cases: Multiple Optimal Solutions vs Infeasible LPP

What happens when an LPP surprises you at the finish line? Sometimes TWO neighboring corner points tie for the exact same best score Z — meaning EVERY point on the straight edge connecting them is ALSO optimal! Other times, two constraints pull in completely opposite directions (like "make at least 8" AND "at most 3") leaving zero overlap — an Infeasible LPP!

MULTIPLE OPTIMALIf two adjacent corners P, Q give the SAME optimal Z, then EVERY point on segment PQ gives that optimal Z (∞ optimal solutions!).

WHY IT HAPPENSThe objective line Z = ax + by runs parallel to boundary edge PQ!

INFEASIBLE LPPConstraints have NO common point (Feasible Region = ∅) → NO feasible solution!

🧠 Memory Trick: Tie at 2 Corners = Infinitely Many Solutions (entire segment PQ!); Empty Overlap = Zero Solutions (Infeasible!)

💡 Simple Kid-Check: Picture sliding your ruler Z = 3x + 9y across a polygon and discovering it aligns flat against the top wall x + 3y = 60 between corners (0, 20) and (15, 15)! Since the ruler hits the whole edge at once, not just the two endpoints, EVERY point along that wall earns the exact same jackpot Z = 180!

LPP Situation	Visual Clue on Graph	Board-Exam Conclusion
Unique Optimal	Best `Z` at 1 single corner	Unique value at that single `(x, y)`
Multiple Optimal	2 adjacent corners tie for best `Z`	Infinitely many optimal solutions along segment joining them!
Infeasible	Shaded arrows point apart (no overlap)	Empty feasible region → No feasible solution!
Unbounded (No Opt)	Open half-plane overlaps feasible region	No maximum / No minimum exists!

🧒 NCERT Example 3 & Example 5 Highlights:
• Multiple Optimal (Ex 3): Maximise Z = 3x + 9y on x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x,y ≥ 0.
Corners: A(0,10)→90, B(5,5)→60 (Min), C(15,15)→180, D(0,20)→180! Both C and D tie at 180 → Max Z = 180 occurs at EVERY point on segment CD!
• Infeasible (Ex 5): x + y ≥ 8 and 3x + 5y ≤ 15 (x, y ≥ 0). Notice x+y≥8 stays outside (8,0)–(0,8) while 3x+5y≤15 stays inside (5,0)–(0,3) → zero overlap → No feasible solution!

When two corner points tie for the maximum (like (15, 15) and (0, 20) both giving Z = 180), writing "There are TWO optimal solutions" is WRONG! You MUST write: "Maximum value 180 occurs at INFINITELY MANY points — every point on the line segment joining (15, 15) and (0, 20)!"

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE Model / PYQ • 1 Marks): Define 'Feasible Region' and 'Optimal Solution' in a Linear Programming Problem.

Feasible Region: The common region determined by all the linear constraints including the non-negativity constraints (x ≥ 0, y ≥ 0) of an LPP.

Optimal Solution: Any point within the feasible region that gives the maximum or minimum value of the objective function Z = ax + by.

Q2 (Kerala HSE March PYQ • 2 Marks): State the Corner-Point Theorem for a bounded feasible region, and explain when an LPP admits infinitely many optimal solutions.

If the feasible region R is bounded, the objective function Z = ax + by attains both a maximum and a minimum value at corner points (vertices) of R.

If two adjacent corner points produce the exact same optimal value of Z, then every point on the line segment joining them gives that same optimal value — resulting in infinitely many optimal solutions.

Q3 (NCERT / Kerala HSE PYQ • 2 Marks): A merchant sells two models of fans A and B. He has space for at most 60 fans and ₹50,000 to invest. Fan A costs ₹2,500 (profit ₹250) and Fan B costs ₹500 (profit ₹75). Formulate this as an LPP to maximise profit.

Let x = number of Fan A and y = number of Fan B.

Maximise: Z = 250x + 75y

Subject to: x + y ≤ 60 (storage), 2500x + 500y ≤ 50000 ⇒ 5x + y ≤ 100 (budget), and x ≥ 0, y ≥ 0.

Q4 (Kerala HSE March PYQ • 3 Marks): Solve graphically: Maximise Z = 3x + 2y subject to x + 2y ≤ 10, 3x + y ≤ 15, x ≥ 0, y ≥ 0.

Boundary lines: x + 2y = 10 through (10,0), (0,5); 3x + y = 15 through (5,0), (0,15). Both shade toward (0,0).

Intersection: 3(10 − 2y) + y = 15 ⇒ 30 − 5y = 15 ⇒ y = 3, x = 4.

Corner points of bounded region: (0,0) → 0, (5,0) → 15, (4,3) → 12 + 6 = 18, (0,5) → 10.

Hence Maximum Z = 18 at (4, 3).

Q5 (Kerala HSE Say / Model PYQ • 3 Marks): Minimise Z = −3x + 4y subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Lines intersect at: subtracting x + 2y = 8 from 3x + 2y = 12 gives 2x = 4 ⇒ x = 2, y = 3.

Bounded vertices: (0,0) → 0, (4,0) → −12, (2,3) → −6 + 12 = 6, (0,4) → 16.

Minimum value of Z is −12 at (4, 0).

Q6 (Kerala HSE March PYQ (High Yield) • 4 Marks): Solve graphically: Minimise Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2, x ≥ 0, y ≥ 0.

Line x + 3y = 3 meets axes at (3,0), (0,1); Line x + y = 2 meets axes at (2,0), (0,2). Both shade AWAY from (0,0) → Unbounded feasible region.

Intersection: (x + 3y) − (x + y) = 3 − 2 ⇒ 2y = 1 ⇒ y = ½, x = 3/2.

Corner points: (3, 0) → 9, (3/2, 1/2) → 9/2 + 5/2 = 7, (0, 2) → 10.

Open Half-Plane Test: Graph dashed 3x + 5y < 7 (passes through (7/3, 0) and (0, 7/5), shades toward (0,0)). It has NO point in common with the feasible region.

Therefore, Minimum Z = 7 at (3/2, 1/2).

Q7 (NCERT Ex 3 / HSE PYQ • 4 Marks): Minimise and Maximise Z = 3x + 9y subject to x + 3y ≤ 60, x + y ≥ 10, x ≤ y, x ≥ 0, y ≥ 0.

Bounded feasible region has 4 corner points:

• A(0, 10) (on x=0, x+y=10) → Z = 90

• B(5, 5) (on x=y, x+y=10) → Z = 15 + 45 = 60 (Minimum!)

• C(15, 15) (on x=y, x+3y=60) → Z = 45 + 135 = 180 (Maximum!)

• D(0, 20) (on x=0, x+3y=60) → Z = 180 (Maximum!)

Conclusion: Minimum Z = 60 at (5, 5); Maximum Z = 180 at every point on the line segment joining (15, 15) and (0, 20).

Q8 (NCERT Ex 4 / HSE Model PYQ • 4 Marks): Determine graphically whether Z = −50x + 20y has a minimum value subject to 2x − y ≥ −5, 3x + y ≥ 3, 2x − 3y ≤ 12, x ≥ 0, y ≥ 0.

The shaded feasible region in Quadrant I is unbounded with corner points (0, 5), (0, 3), (1, 0), and (6, 0).

Corner values: Z(0,5) = 100, Z(0,3) = 60, Z(1,0) = −50, Z(6,0) = −300.

Open Half-Plane Check: Graph dashed −50x + 20y < −300 (i.e. −5x + 2y < −30). Test feasible point (10, 4): −500 + 80 = −420 < −300 is TRUE!

Since the open half-plane has common points with the feasible region, Z has NO minimum value.