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Graphical max/min min:z=9x+40y

Short answer

The feasible polygon has corners (2,3), (5,6), (5,8), (2,5). Evaluating z = 9x + 40y at each yields maximum 365 at (5,8) and minimum 138 at (2,3).

Graphically maximize and minimize z = 9x + 40y subject to the constraints y − x ≥ 1, y − x ≤ 3, 2 ≤ x ≤ 5.

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) −x + y ≥ 1−x + y = 1(−1, 0)(0, 1)away from the origin
(2) −x + y ≤ 3−x + y = 3(−3, 0)(0, 3)towards the origin
(3) x ≥ 2x = 2(2, 0)none (parallel to y-axis)away from the origin
(4) x ≤ 5x = 5(5, 0)none (parallel to y-axis)towards the origin

Also x ≥ 0 and y ≥ 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, ….

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

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 = 9x + 40y
A(2, 3)(1), (3)138
B(5, 6)(1), (4)285
C(5, 8)(2), (4)365
D(2, 5)(2), (3)218

Step 4: Conclusion. The largest value is 365 at C(5, 8) and the smallest value is 138 at A(2, 3).

Answer: Maximum Z = 365 at x = 5, y = 8; Minimum Z = 138 at x = 2, y = 3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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