i) Formulate the problem using a standard linear programming simplex algorithm format.

Here's the solution to the linear programming problem: i) Formulate the problem using a standard linear programming simplex algorithm format. Let x_P, x_Q, and x_R be the number of units produced for products P, Q, and R, respectively. Objective Function (Maximize Profit Z): Maximize Z = 200x_P + 200x_Q + 500x_R Subject to the following constraints: • Molding time: 2x_P + 1.5x_Q + 2.5x_R 400 • Painting time: 1x_P + 1x_Q + 1x_R 200 • Assembling time: 1.5x_P + 1x_Q + 1.5x_R 300 Non-negativity constraints: x_P, x_Q, x_R 0 To convert this to the standard linear programming simplex algorithm format, we introduce slack variables (s_1, s_2, s_3) for each inequality constraint to turn them into equalities, and express the objective function in a form suitable for the simplex tableau. Standard Simplex Format: Maximize Z = 200x_P + 200x_Q + 500x_R + 0s_1 + 0s_2 + 0s_3 Subject to: 2x_P + 1.5x_Q + 2.5x_R + s_1 = 400 x_P + x_Q + x_R + s_2 = 200 1.5x_P + x_Q + 1.5x_R + s_3 = 300 x_P, x_Q, x_R, s_1, s_2, s_3 0 And for the simplex tableau, the objective function is typically written as: -200x_P - 200x_Q - 500x_R + Z = 0 ii) Determine the quantity of each product that should be produced so as to maximize profit. To find the optimal quantities, we solve the linear programming problem. Using a linear programming solver, the optimal production quantities are: • x_P = 0 units • x_Q = 0 units • x_R = 160 units iii) Determine the profit realizable at the optimal level of production. Substitute the optimal quantities into the objective function: Z = 200(0) + 200(0) + 500(160) Z = 0 + 0 + 80000 Z = N80,000 The maximum profit realizable is N80,000. iv) Can all the products be produced at this optimal level of production? No, at the optimal level of production, only product R is produced. Products P and Q are not produced. v) Is there any unutilized time in any of the departments? Which and by what quantity? We check the usage of each resource with the optimal production quantities (x_P=0, x_Q=0, x_R=160): • Molding: 2(0) + 1.5(0) + 2.5(160) = 400 hours. Available: 400 hours. Unutilized time: 400 - 400 = 0 hours. • Painting: 1(0) + 1(0) + 1(160) = 160 hours. Available: 200 hours. Unutilized time: 200 - 160 = 40 hours. • Assembling: 1.5(0) + 1(0) + 1.5(160) = 240 hours. Available: 300 hours. Unutilized time: 300 - 240 = 60 hours. Therefore, there is unutilized time in the Painting department (40 hours) and the Assembling department (60 hours). That's 2 down. 3 left today — send the next one.