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Graphical method: max and min

Short answer

Plot the constraints, identify corner points of the feasible region, and evaluate Z = 50x₁ + 80x₂. Maximum Z = 1480 at (20, 6); Minimum Z = 0 at (0, 0).

Find the maximum and minimum values of the function Z = 50x₁ + 80x₂ subject to x₁ + 2x₂ ≤ 32, 3x₁ + 4x₂ ≤ 84, x₁, x₂ ≥ 0 by the graphical method.

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) x₁ + 2x₂ ≤ 32x₁ + 2x₂ = 32(32, 0)(0, 16)towards the origin
(2) 3x₁ + 4x₂ ≤ 843x₁ + 4x₂ = 84(28, 0)(0, 21)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, ….

  18|                                            
    |                                            
    |D                                           
    |....                                        
    |.......                                     
    |..........                                  
    |.............                               
    |................                            
    |...................                         
    |......................                      
    |.........................                   
    |...........................C                
    |..............................              
    |................................            
    |..................................          
    |....................................        
   0|A.....................................B     
    +--------------------------------------------
     0                                         32

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 = 50x₁ + 80x₂
A(0, 0)x₁ = 0, x₂ = 00
B(28, 0)(2), x₂ = 01400
C(20, 6)(1), (2)1480
D(0, 16)(1), x₁ = 01280

Step 4: Conclusion. The largest value is 1480 at C(20, 6) and the smallest value is 0 at A(0, 0).

Answer: Maximum Z = 1480 at x₁ = 20, x₂ = 6; Minimum Z = 0 at x₁ = 0, x₂ = 0.

✓ Answer checked by Claude Opus 5.5 (high effort)

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