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 $Z$): $\text{Maximize } Z = 200x_P + 200x_Q + 500x_R$

Subject to the following constraints:

• Molding time: $2x_P + 1.5x_Q + 2.5x_R \le 400$
• Painting time: $1x_P + 1x_Q + 1x_R \le 200$
• Assembling time: $1.5x_P + 1x_Q + 1.5x_R \le 300$

Non-negativity constraints: $x_P, x_Q, x_R \ge 0$

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: $\text{Maximize } Z = 200x_P + 200x_Q + 500x_R + 0s_1 + 0s_2 + 0s_3$ Subject to: $2x_P + 1.5x_Q + 2.5x_R + s_1 = 400$ $x_P + x_Q + x_R + s_2 = 200$ $1.5x_P + x_Q + 1.5x_R + s_3 = 300$ $x_P, x_Q, x_R, s_1, s_2, s_3 \ge 0$ 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:

• $x_P = \mathbf{0}$ units
• $x_Q = \mathbf{0}$ units
• $x_R = \mathbf{160}$ units

iii) Determine the profit realizable at the optimal level of production. Substitute the optimal quantities into the objective function: $Z = 200(0) + 200(0) + 500(160)$ $Z = 0 + 0 + 80000$ $Z = \mathbf{N80,000}$ 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: $400 - 400 = \mathbf{0}$ hours.
• Painting: $1(0) + 1(0) + 1(160) = 160$ hours. Available: 200 hours. Unutilized time: $200 - 160 = \mathbf{40}$ hours.
• Assembling: $1.5(0) + 1(0) + 1.5(160) = 240$ hours. Available: 300 hours. Unutilized time: $300 - 240 = \mathbf{60}$ hours.

Therefore, there is unutilized time in the Painting department (40 hours) and the Assembling department (60 hours).