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Graphical max/min opt:Z=4x₁+5x₂

Short answer

Draw x₁+x₂≤10, -x₁+x₂≥2, x₁,x₂≥0. Corner points: (0,2) Z=10, (4,6) Z=46, (0,10) Z=50. Maximum Z=50 at x₁=0, x₂=10; minimum Z=10 at x₁=0, x₂=2.

Find the maximum and minimum values of the following linear programming problem by using the graphical method: Optimize Z = 4x₁ + 5x₂ subject to x₁ + x₂ ≤ 10, −x₁ + x₂ ≥ 2, x₁, x₂ ≥ 0.

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 + x₂ ≤ 10x₁ + x₂ = 10(10, 0)(0, 10)towards the origin
(2) −x₁ + x₂ ≥ 2−x₁ + x₂ = 2(−2, 0)(0, 2)away from 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, ….

  12|                                            
    |                                            
    |                                            
    |C..                                         
    |.........                                   
    |................                            
    |......................                      
    |............................                
    |..................................B         
    |............................                
    |......................                      
    |................                            
    |.........                                   
    |A..                                         
    |                                            
    |                                            
   0|                                            
    +--------------------------------------------
     0                                          5

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 = 4x₁ + 5x₂
A(0, 2)(2), x₁ = 010
B(4, 6)(1), (2)46
C(0, 10)(1), x₁ = 050

Step 4: Conclusion. The largest value is 50 at C(0, 10) and the smallest value is 10 at A(0, 2).

Answer: Maximum Z = 50 at x₁ = 0, x₂ = 10; Minimum Z = 10 at x₁ = 0, x₂ = 2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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