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Graphical method max:Z=x₁+1.5x₂

Short answer

Plot the three constraints and find corner points: (0,0), (70,0), (60,20), (40,40), (0,60). Evaluating Z = x₁ + 1.5x₂ gives maximum 100 at x₁ = 40, x₂ = 40.

Maximize Z = x₁ + 1.5x₂ subject to the constraints 2x₁ + 2x₂ ≤ 160, x₁ + 2x₂ ≤ 120, 4x₁ + 2x₂ ≤ 280, x₁ ≥ 0 and x₂ ≥ 0 graphically.

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).

ConstraintLineMeets x-axisMeets y-axisAllowed side
(1) 2x₁ + 2x₂ ≤ 1602x₁ + 2x₂ = 160(80, 0)(0, 80)towards the origin
(2) x₁ + 2x₂ ≤ 120x₁ + 2x₂ = 120(120, 0)(0, 60)towards the origin
(3) 4x₁ + 2x₂ ≤ 2804x₁ + 2x₂ = 280(70, 0)(0, 140)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, ….

  68|                                            
    |                                            
    |E                                           
    |......                                      
    |..........                                  
    |...............                             
    |....................                        
    |......................D                     
    |..........................                  
    |............................                
    |..............................              
    |................................ C          
    |..................................          
    |...................................         
    |....................................        
    |.....................................       
   0|A.....................................B     
    +--------------------------------------------
     0                                         79

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

PointCoordinatesLines meeting thereZ = x₁ + 1.5x₂
A(0, 0)x₁ = 0, x₂ = 00
B(70, 0)(3), x₂ = 070
C(60, 20)(1), (3)90
D(40, 40)(1), (2)100
E(0, 60)(2), x₁ = 090

Step 4: Conclusion. The largest value of Z is 100, at D(40, 40).

Answer: Maximum Z = 100 at x₁ = 40, x₂ = 40.

✓ Answer checked by Claude Opus 5.5 (high effort)

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