---
title: "Graphical method max:Z=7x₁+10x₂"
subject: "Mathematics III"
chapter: 5
key: "semester_3/mathematics_iii:graphical-method-max-z-7x-10x-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-7x-10x-two
site: BCT Engineering
---

# Graphical method max:Z=7x₁+10x₂

Mathematics III · Chapter 5: Linear Programming

- Asked in: 2083 Baishakh, 5 marks
- Video: [Graphical Method for LP Problem |  Graphical + Corner Point Method | Operations Research | MBA](https://www.youtube.com/watch?v=fWHzKpqZctE) — KN Prasad Academy (NITian)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/graphical-method-max-z-7x-10x-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-5

## Question

Applying the graphical method, maximize Z = 7x₁ + 10x₂ subject to the constraints 3x₁ + x₂ ≤ 9, x₁ + x₂ ≤ 8, x₁, x₂ ≥ 0.

## Short answer

Plot lines 3x₁+x₂=9, x₁+x₂=8; feasible region corners: (0,0) Z=0, (3,0) Z=21, (0.5,7.5) Z=78.5, (0,8) Z=80. Maximum Z=80 at x₁=0, x₂=8.

## Answer

**Step 1: Draw the constraint lines.** Replace each inequality by an equation, find the intercepts and decide which side of the line is allowed (test the origin).

| Constraint | Line | Meets x-axis | Meets y-axis | Allowed side |
|---|---|---|---|---|
| (1) 3x₁ + x₂ ≤ 9 | 3x₁ + x₂ = 9 | (3, 0) | (0, 9) | towards the origin |
| (2) x₁ + x₂ ≤ 8 | x₁ + x₂ = 8 | (8, 0) | (0, 8) | towards the origin |

Also x₁ ≥ 0 and x₂ ≥ 0, so the region lies in the first quadrant.

**Step 2: Feasible region.** The region satisfying all constraints is the shaded polygon below (dots), with corner points marked A, B, C, ….

```text
   9|                                            
    |                                            
    |D.                                          
    |.....C.                                     
    |.........                                   
    |...........                                 
    |.............                               
    |...............                             
    |.................                           
    |...................                         
    |.....................                       
    |.......................                     
    |.........................                   
    |...........................                 
    |.............................               
    |...............................             
   0|A...............................B           
    +--------------------------------------------
     0                                          4
```

**Step 3: Corner points and the objective value.** The optimum of a linear function over a convex polygon occurs at a corner point.

| Point | Coordinates | Lines meeting there | Z = 7x₁ + 10x₂ |
|---|---|---|---|
| A | (0, 0) | x₁ = 0, x₂ = 0 | 0 |
| B | (3, 0) | (1), x₂ = 0 | 21 |
| C | (1/2, 15/2) | (1), (2) | 157/2 |
| D | (0, 8) | (2), x₁ = 0 | 80 |

**Step 4: Conclusion.** The largest value of Z is 80, at D(0, 8).

**Answer:** Maximum Z = 80 at x₁ = 0, x₂ = 8.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the intercepts of x1 + x2 = 8?
   A. (8, 0) and (0, 8)
   B. (8, 0) and (0, 4)
   C. (4, 0) and (0, 8)
   D. (9, 0) and (0, 8)
   Answer: A. Equal coefficients give equal intercepts. (Key point: Intercepts)

2. Where do the lines meet?
   A. (2, 3)
   B. (1, 7)
   C. (1/2, 15/2)
   D. (3, 0)
   Answer: C. Subtract: 2x1 = 1, x1 = 1/2, x2 = 15/2. (Key point: Intersection C)

3. Why is the maximum at D(0, 8) and not C?
   A. Z = 80 at D beats 157/2 = 78.5 at C
   B. C is infeasible
   C. Z at C is 44
   D. D lies on the first line
   Answer: A. 157/2 < 80. (Key point: Comparison)

4. What is the maximum Z?
   A. 72 at (0, 7.2)
   B. 21 at (3, 0)
   C. 80 at x1 = 0, x2 = 8
   D. 78.5 at (1/2, 15/2)
   Answer: C. 7(0) + 10(8) = 80. (Key point: Answer)

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