(a) The decision variables are:

• x = number of journeys made by Lorry X
• y = number of journeys made by Lorry Y

(b) The linear programming problem is formulated as follows:

Minimize Z = 20000x + 15000y (total running costs in GHS)

Subject to: 10x + 3y ≥ 200 (minimum delivery constraint for Product A) 4x + 5y ≥ 150 (minimum delivery constraint for Product B) x ≥ 2 (minimum journeys for Lorry X) y ≥ 2 (minimum journeys for Lorry Y) x ≥ 0, y ≥ 0 (non-negativity constraints)

(c) To display the LPM on a graph and shade the critical region (feasible region):

The graph has x (journeys by Lorry X) on the horizontal axis and y (journeys by Lorry Y) on the vertical axis.

• Plot the line 10x + 3y = 200: intercepts at (20, 0) and (0, 66.67).
• Plot the line 4x + 5y = 150: intercepts at (37.5, 0) and (0, 30).
• Plot the lines x = 2 (vertical line) and y = 2 (horizontal line).

The feasible region is the area where all inequalities are satisfied: to the right of x = 2, above y = 2, above the line 10x + 3y = 200, and above the line 4x + 5y = 150. This region is unbounded in the top-right direction and has corner points at approximately (2, 60), (14.47, 18.42), and (35, 2).

(d) To solve the LPM using the graphical approach:

First, identify the corner points of the feasible region:

1. Intersection of x = 2 and 10x + 3y = 200: 10(2) + 3y = 200 → 20 + 3y = 200 → 3y = 180 → y = 60 Point: (2, 60) Cost Z = 20000(2) + 15000(60) = 40000 + 900000 = 940000 GHS
2. Intersection of 10x + 3y = 200 and 4x + 5y = 150: Solve the system: Multiply the first equation by 5 and the second by 3: 50x + 15y = 1000 12x + 15y = 450 Subtract: 38x = 550 → x = 550/38 = 275/19 ≈ 14.47 Substitute into 4x + 5y = 150: 4(275/19) + 5y = 150 → 1100/19 + 5y = 150 → 5y = 150 – 1100/19 = (2850 – 1100)/19 = 1750/19 y = (1750/19)/5 = 1750/95 = 350/19 ≈ 18.42 Point: (275/19, 350/19) Cost Z = 20000(275/19) + 15000(350/19) = (5500000 + 5250000)/19 = 10750000/19 ≈ 565789 GHS
3. Intersection of y = 2 and 4x + 5y = 150: 4x + 5(2) = 150 → 4x + 10 = 150 → 4x = 140 → x = 35 Point: (35, 2) Cost Z = 20000(35) + 15000(2) = 700000 + 30000 = 730000 GHS

Compare the costs at the corner points: 940000 at (2, 60), 565789 at (14.47, 18.42), and 730000 at (35, 2). The minimum cost is approximately 565789 GHS at x ≈ 14.47 journeys by Lorry X and y ≈ 18.42 journeys by Lorry Y.

To arrive at the solution: The graphical method involves plotting the constraints to form the feasible region, identifying the corner points where constraint lines intersect, and evaluating the objective function at each corner point. The point yielding the minimum value of the objective function is the optimal solution. Since the objective is minimization, the optimal occurs at the corner closest to the origin in the direction opposite to the gradient of the objective function (or by shifting the iso-cost line towards lower costs until it last touches the feasible region at the intersection point).