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Graphical method max:Z=4x₁+5x₂

Short answer

The feasible region is a quadrilateral with vertices (0,0), (7.5,0), (5,3), (0,5). The objective Z = 4x₁ + 5x₂ attains its maximum 35 at x₁ = 5, x₂ = 3.

Maximize Z = 4x₁ + 5x₂ subject to constraints 2x₁ + 5x₂ ≤ 25, 6x₁ + 5x₂ ≤ 45, 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 + 5x₂ ≤ 252x₁ + 5x₂ = 25(25/2, 0)(0, 5)towards the origin
(2) 6x₁ + 5x₂ ≤ 456x₁ + 5x₂ = 45(15/2, 0)(0, 9)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, ….

   6|                                            
    |                                            
    |                                            
    |D.                                          
    |......                                      
    |...........                                 
    |...............                             
    |....................                        
    |........................C                   
    |..........................                  
    |...........................                 
    |.............................               
    |..............................              
    |................................            
    |.................................           
    |...................................         
   0|A...................................B       
    +--------------------------------------------
     0                                          9

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, 0)x₁ = 0, x₂ = 00
B(15/2, 0)(2), x₂ = 030
C(5, 3)(1), (2)35
D(0, 5)(1), x₁ = 025

Step 4: Conclusion. The largest value of Z is 35, at C(5, 3).

Answer: Maximum Z = 35 at x₁ = 5, x₂ = 3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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