A complete guide to turning a rice-trading story into a math model, then finding the highest possible sales with confidence.
AMI | Global Mathematics Education | www.ami.sch.id
A rice trader wants to make blended rice by mixing type A rice and type B rice. The first blend consists of 4 kg of type A and 8 kg of type B, while the second blend consists of 8 kg of type A and 10 kg of type B. The available stock of types A and B is 80 tonnes and 106 tonnes respectively. If the selling price is Rp60,000.00 for the first blend and Rp80,000.00 for the second, the maximum sales that can be obtained is ...
A. Rp1,200,000,000.00
B. Rp920,000,000.00
C. Rp840,000,000.00
D. Rp800,000,000.00
E. Rp795,000,000.00
Study Map
| Ch. | Section | What you will master |
|---|---|---|
| 1 | Dissecting the Problem | Reading the story, Given and To Find table, 5-step flow |
| 2 | Math Foundations | Terms, story-to-inequality dictionary, basic formulas |
| 3 | Core Concepts | General form, types of cases, corner-point rule, choosing a method |
| 4 | Complete Solution | Four solution methods, verification, and the answer |
| 5 | Quick Formulas and Fast Tricks | Thousands trick, quick intersection, slope check, guess-and-check |
| 6 | Watch Out for Traps | Wrong vs. right, champion habits, did-you-know fact |
| 7 | Graded Practice | Four problems from easy to HOTS, with answer key |
| 8 | Formula Card | One-page cheat sheet, six colorful boxes |
CHAPTER 1 DISSECTING THE PROBLEM
A rice trader wants to make blended rice by mixing type A rice and type B rice. The first blend consists of 4 kg of type A and 8 kg of type B, while the second blend consists of 8 kg of type A and 10 kg of type B. The available stock of types A and B is 80 tonnes and 106 tonnes respectively. If the selling price is Rp60,000.00 for the first blend and Rp80,000.00 for the second, the maximum sales that can be obtained is ...
A. Rp1,200,000,000.00 B. Rp920,000,000.00 C. Rp840,000,000.00 D. Rp800,000,000.00 E. Rp795,000,000.00
Illustration of the problem
Given and To Find
| Given | Meaning in math language |
|---|---|
| Blend I: 4 kg A and 8 kg B, price Rp60,000 | Each Blend I uses 4 kg of A, 8 kg of B, and earns Rp60,000 |
| Blend II: 8 kg A and 10 kg B, price Rp80,000 | Each Blend II uses 8 kg of A, 10 kg of B, and earns Rp80,000 |
| Stock of rice A: 80 tonnes | Use of A is at most 80,000 kg (1 tonne = 1,000 kg) |
| Stock of rice B: 106 tonnes | Use of B is at most 106,000 kg |
| To find | The maximum sales (the maximum value of the objective function) |
The 5-step flow to solve it
CHAPTER 2 MATH FOUNDATIONS
Table of terms
| Term | Meaning | Example from the problem |
|---|---|---|
| Decision variable | The quantity we want to find | x and y = number of Blend I and Blend II |
| Constraint | An inequality from a resource limit | x + 2y ≤ 20 |
| Objective function | The quantity to maximize or minimize | Z = 60x + 80y |
| Feasible region | All points that satisfy every constraint | A quadrilateral with 4 corner points |
| Corner point | A vertex of the feasible region | (0, 0), (13.25, 0), (2, 9), (0, 10) |
| Isoprofit line | A line parallel to the objective, slid around | 60x + 80y = k |
| Optimal value | The largest or smallest value of Z | Maximum Z = 840 |
Translation dictionary: sentences become symbols
| Sentence or data | Written as | In this problem |
|---|---|---|
| Available, at most | Sign ≤ | Rice A used ≤ 80 tonnes |
| At least | Sign ≥ | Not in this problem |
| The number of blends cannot be negative | x ≥ 0 and y ≥ 0 | x ≥ 0, y ≥ 0 |
| Selling price Rp60,000 per Blend I | Z increases by 60 per unit of x | Z = 60x + 80y (million rupiah) |
| 1 tonne | 1,000 kg | 80 tonnes = 80,000 kg |
| In thousands | Divide every number by 1,000 | 4x + 8y ≤ 80 |
Basic properties and formulas you need
| Formula | Used for | Example in this problem |
|---|---|---|
| Z = px + qy | Objective function | Z = 60x + 80y |
| a1x + b1y ≤ c1 | Constraint | 4x + 5y ≤ 53 |
| m = −ab | Slope of the line ax + by = c | m1 = −12 = −0.5 |
| x = c1b2 − c2b1a1b2 − a2b1 | Intersection of two lines (x value) | x = 20·5 − 53·21·5 − 4·2 = −6−3 = 2 |
| y = a1c2 − a2c1a1b2 − a2b1 | Intersection of two lines (y value) | y = 1·53 − 4·201·5 − 4·2 = −27−3 = 9 |
Make the units consistent before building the model. Here the blends are in kg but the stock is in tonnes, so convert 1 tonne to 1,000 kg first.
CHAPTER 3 CORE CONCEPTS
Linear programming is a method for finding the maximum or minimum value of a linear function (the objective function) whose variables are limited by a system of linear inequalities (the constraints).
General form of a linear programming model
| Part of the model | Form | In this problem |
|---|---|---|
| Objective function | Z = px + qy | Z = 60x + 80y |
| Constraints | a1x + b1y ≤ c1, a2x + b2y ≤ c2 | x + 2y ≤ 20 and 4x + 5y ≤ 53 |
| Non-negativity | x ≥ 0, y ≥ 0 | x ≥ 0, y ≥ 0 |
Types of feasible regions
| Problem type | Constraint sign | Optimum is usually at |
|---|---|---|
| Maximizing (this problem) | Constraints ≤ (upper limits on resources) | The corner point farthest from the origin |
| Minimizing | Constraints ≥ (minimum requirements) | The corner point closest to the origin |
If the feasible region is bounded, the maximum and minimum of the objective function always occur at one of the corner points. So it is enough to compute Z at the corner points.
Comparing methods: when to use which?
| Method | Best when | Strength | Be careful |
|---|---|---|---|
| Corner-point test | The feasible region is bounded | Always right, easy to check | Do not miss the intersection of two lines |
| Isoprofit line | You have a graph or want to see the direction | Shows the optimum clearly | Needs an accurate drawing |
| Slope comparison | Two constraints and positive Z | Quickly guesses the optimal point | Only picks the point; still compute the value |
| Upper bound (multipliers) | Multiple choice, need a proof | Proves no larger value is possible | Multipliers must be positive |
CHAPTER 4 COMPLETE SOLUTION
Building the mathematical model (used by every method)
| Material | Blend I | Blend II | Stock |
|---|---|---|---|
| Rice A (kg) | 4 | 8 | 80,000 |
| Rice B (kg) | 8 | 10 | 106,000 |
| Selling price (Rp) | 60,000 | 80,000 | - |
| Step | Work |
|---|---|
| 1. Define (in thousands of blends) | x = number of Blend I, y = number of Blend II |
| 2. Constraint for rice A | 4x + 8y ≤ 80, simplify to x + 2y ≤ 20 |
| 3. Constraint for rice B | 8x + 10y ≤ 106, simplify to 4x + 5y ≤ 53 |
| 4. Non-negativity | x ≥ 0 and y ≥ 0 |
| 5. Objective function (million rupiah) | Z = 60x + 80y, because 1 thousand Blend I is worth 1,000 × Rp60,000 = Rp60 million |
Method 1: Corner-Point Test
| Step | Work |
|---|---|
| 1. Axis intercepts of x + 2y = 20 | (20, 0) and (0, 10) |
| 2. Axis intercepts of 4x + 5y = 53 | (13.25, 0) and (0, 10.6) |
| 3. Intersection of the two lines (determinants) | x = 20·5 − 53·21·5 − 4·2 = −6−3 = 2 y = 1·53 − 4·201·5 − 4·2 = −27−3 = 9 The intersection is (2, 9) |
| 4. Keep the feasible points | (20, 0) is infeasible because 4·20 = 80 > 53. (0, 10.6) is infeasible because 2·10.6 = 21.2 > 20. Corner points: (0, 0), (13.25, 0), (2, 9), (0, 10) |
The feasible region and Z at the corner points
The feasible region is shaded yellow. The orange point (2, 9) is where the two constraints meet.
| Corner point | Z = 60x + 80y | Z (million rupiah) |
|---|---|---|
| (0, 0) | 60·0 + 80·0 | 0 |
| (13.25, 0) | 60·13.25 + 80·0 | 795 |
| (0, 10) | 60·0 + 80·10 | 800 |
| (2, 9) | 60·2 + 80·9 = 120 + 720 | 840 (largest) |
Method 2: Isoprofit Line
| Step | Work |
|---|---|
| 1. Write the isoprofit line | 60x + 80y = k, or 3x + 4y = k |
| 2. Draw several parallel lines | For example k = 300, 600, and 840 (see the graph) |
| 3. Slide to the upper right | As long as the line still touches the feasible region, k keeps increasing |
| 4. The last point touched | Point (2, 9), so the maximum k = 60·2 + 80·9 = 840 |
Method 3: Slope Comparison
| Step | Work |
|---|---|
| 1. Slope of the rice A constraint (x + 2y = 20) | m1 = −12 = −0.5 |
| 2. Slope of the rice B constraint (4x + 5y = 53) | m2 = −45 = −0.8 |
| 3. Slope of the objective (60x + 80y = k) | mZ = −6080 = −0.75 |
| 4. Compare | −0.8 < −0.75 < −0.5 The slope of Z lies between the slopes of the two constraints |
| 5. Conclusion | The optimum is at the intersection of the two constraints, (2, 9); Z = 840 |
Method 4: Upper Bound with Multipliers
| Step | Work |
|---|---|
| 1. Find multipliers λ1, λ2 so that λ1(x + 2y) + λ2(4x + 5y) = 60x + 80y | λ1 + 4λ2 = 60, 2λ1 + 5λ2 = 80 ⇒ λ1 = 203, λ2 = 403 |
| 2. Write Z as a combination of the constraints | 60x + 80y = 203(x + 2y) + 403(4x + 5y) |
| 3. Since the constraints are ≤ and the multipliers are positive | Z ≤ 203·20 + 403·53 = 400 + 21203 = 840 |
| 4. Check that the bound is reached | At (2, 9) both constraints are tight (20 and 53), so Z = 840 is actually attained |
Verification: test the optimal point (2,000, 9,000)
| What is tested | Calculation | Result |
|---|---|---|
| Rice A does not exceed 80 tonnes | 4(2) + 8(9) = 8 + 72 = 80 ≤ 80 (thousand kg) | Match |
| Rice B does not exceed 106 tonnes | 8(2) + 10(9) = 16 + 90 = 106 ≤ 106 | Match |
| x ≥ 0 and y ≥ 0 | 2 ≥ 0 and 9 ≥ 0 | Match |
| Sales value | 60(2) + 80(9) = 840 million rupiah | Match |
| No other corner point is larger | 795 and 800 are smaller than 840 | Match |
Achieved by making 2,000 Blend I and 9,000 Blend II
CHAPTER 5 QUICK FORMULAS AND FAST TRICKS
Three quick tricks
Thousands: dividing both sides of an inequality by a positive number (1,000) does not change its sign, and Z in million rupiah keeps the numbers short.
Quick intersection: from a1x + b1y = c1 and a2x + b2y = c2, multiply the first equation by b2 and the second by b1, then subtract. The result:
x = c1b2 − c2b1a1b2 − a2b1
Condition: the denominator a1b2 − a2b1 must not be zero (the lines must not be parallel). The y value is found the same way.
Trick 3: Check the Slopes to Predict the Optimal Point
| Slope of Z compared with the constraint slopes (absolute values) | Optimal point (maximize, constraints ≤) |
|---|---|
| Flatter than both (|mZ| < 0.5) | On the Y axis: (0, 10) |
| Between them (0.5 < |mZ| < 0.8); this problem: 0.75 | At the intersection of the two lines: (2, 9) |
| Steeper than both (|mZ| > 0.8) | On the X axis: (13.25, 0) |
Reason: the isoprofit line slides in parallel, and the last corner it touches depends on its steepness compared with the sides of the feasible region. This rule applies to the model in this problem (maximize, constraints ≤, a and b positive).
Backup Strategy: Guess-and-Check the Options
| Option | Where the number comes from | Verdict |
|---|---|---|
| A. Rp1,200 million | = 60 × 20, using point (20, 0) which violates constraint B (4·20 = 80 > 53) | Rejected |
| B. Rp920 million | Exceeds the 840 million upper bound from Method 4, so it is impossible | Rejected |
| C. Rp840 million | Value at the feasible point (2, 9), equal to the upper bound | Correct |
| D. Rp800 million | Value at point (0, 10) only; a feasible point gives more (840) | Rejected |
| E. Rp795 million | Value at point (13.25, 0) only; a feasible point gives more | Rejected |
Every point you use must be checked against all constraints. The axis intercept of a single line often looks tempting but may violate another constraint.
CHAPTER 6 WATCH OUT FOR TRAPS!
Wrong vs. Right: the 6 most common mistakes
| No | WRONG | RIGHT | Why |
|---|---|---|---|
| 1 | Using 80 and 106 without converting units | 80 tonnes = 80,000 kg and 106 tonnes = 106,000 kg | Blends are in kg, stock is in tonnes |
| 2 | Treating Rp60,000 as the price per kg | Rp60,000 is the price of one blend | The problem gives the selling price of the blended rice |
| 3 | Taking point (20, 0) without checking | Test all constraints: 4·20 = 80 > 53, infeasible | This is where option A (1,200 million) comes from |
| 4 | Using rice A's numbers (4 and 8) in constraint B | Rice B uses 8 and 10: 8x + 10y ≤ 106 | Each type of rice has its own row |
| 5 | Computing only the axis points (answering D or E) | Also compute the intersection (2, 9) | The optimum is often at an intersection; 800 and 795 are smaller than 840 |
| 6 | mZ = 6080 = 0.75 | mZ = −6080 = −0.75 | The line 60x + 80y = k falls to the right, so its slope is negative |
- Write the units first, then build the data table.
- Name the variables with their units, for example x in thousands of blends.
- Check every corner point against all constraints before computing Z.
- Compare your result with the options, then trace where the other options come from.
Linear programming is used to schedule flights, mix ingredients in factories, and plan delivery routes. Huge problems with thousands of variables are solved by computers.
The simplex method, a very popular way to solve linear programs, was developed by George Dantzig in 1947. Its idea is similar to what you did here: move from one corner point to a better one.
CHAPTER 7 GRADED PRACTICE
Work through them in order. Every problem follows the same pattern as today's problem and has a whole-number answer. HOTS stands for Higher-Order Thinking Skills, meaning a harder problem that needs deeper reasoning.
Answer key and short solutions
| No | Answer | Short solution |
|---|---|---|
| 1 | Rp180,000 | Constraints: x + 3y ≤ 30 and 2x + y ≤ 20; Z = 10x + 15y (thousand rupiah). x = 30·1 − 20·31·1 − 2·3 = 6, y = 1·20 − 2·301·1 − 2·3 = 8 Corner points: (0, 0) = 0; (10, 0) = 100; (0, 10) = 150; (6, 8) = 60 + 120 = 180. |
| 2 | Rp600,000 | Constraints: 2x + y ≤ 40 and x + 3y ≤ 45; Z = 20x + 30y. x = 40·3 − 45·12·3 − 1·1 = 15, y = 2·45 − 1·402·3 − 1·1 = 10 Corner points: (20, 0) = 400; (0, 15) = 450; (15, 10) = 300 + 300 = 600. |
| 3 | Rp420,000,000 | In thousands of blends: 2x + 4y ≤ 24 becomes x + 2y ≤ 12; 3x + 2y ≤ 28; Z = 40x + 50y (million). x = 12·2 − 28·21·2 − 3·2 = 8, y = 1·28 − 3·121·2 − 3·2 = 2 Corner points: (0, 6) = 300; (28/3, 0) ≈ 373.3; (8, 2) = 320 + 100 = 420. |
| 4 | Rp75,000 ≤ q ≤ Rp120,000 | (2, 9) stays optimal when the slope of Z (−60/q) lies between −4/5 and −1/2 (with q in thousand rupiah): −45 ≤ −60q ≤ −12 ⇒ 75 ≤ q ≤ 120 Check: q = 100 gives Z = 60·2 + 100·9 = 1,020 million at (2, 9), larger than 1,000 at (0, 10). |
CHAPTER 8 FORMULA CARD
A quick cheat sheet. Save it or print it, then stick it on your study desk.
Add x ≥ 0 and y ≥ 0. Constraints come from resource limits.
Condition: the denominator is not zero.
The optimum is at a corner point of the feasible region. Compute Z at each feasible corner point, then take the largest (maximum) or smallest (minimum).
If the slope of Z lies between the slopes of the two constraints, the optimum is at their intersection.
Available or at most: sign ≤. At least: sign ≥. 1 tonne = 1,000 kg. The price per blend goes into the objective function.
Option C: Rp840,000,000.00. Optimal point (2, 9): 2,000 Blend I and 9,000 Blend II.
- Make the units consistent, then write the data table.
- Build the constraints, non-negativity, and objective function.
- Find the corner points of the feasible region (do not forget the intersection of two lines).
- Compute Z at every feasible corner point.
- Verify the result against the constraints, then choose the answer.