Linear-Programming Product Mix for Maximum Profit: 23-Chem-B3 Simulation, Modelling, and Optimization, December 2014, Q7

Question 7: Linear-Programming Product Mix for Maximum Profit (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two products with resource requirements and a minimum-labour constraint:

Requirement per year	Product A	Product B
Chemical 1 (180 units available/yr)	18 units/t	3 units/t
Chemical 2 (160 units available/yr)	40 units/t	2 units/t
Labour time (at least 120 units/yr consumed)	24 units/t	4 units/t
Selling price	7 MU/t	4 MU/t

Find. (i) the profit-maximizing $(x_1,x_2)$; (ii) the feasibility region with the optimum marked.

Approach. Formulate the linear program, enumerate the corner points of the feasible polygon, and evaluate the objective at each (the optimum of an LP is attained at a vertex).

LP formulation. Maximize profit $Z = 7x_1 + 4x_2$ subject to $$18x_1 + 3x_2 \le 180,\quad 40x_1 + 2x_2 \le 160,\quad 24x_1 + 4x_2 \ge 120,\quad x_1,x_2 \ge 0.$$
Corner points. The binding-constraint intersections that are feasible are $$(0,30),\ (3.57,\ 8.57),\ (1.43,\ 51.43),\ (0,60).$$ (Note the labour minimum makes an all-$A$ operation infeasible: producing only A cannot consume 120 labour units within the Chemical-2 limit.)
Evaluate the objective. $$Z(0,30)=120,\quad Z(3.57,8.57)=59.3,\quad Z(1.43,51.43)=215.7,\quad Z(0,60)=240.$$
Optimum. The largest profit is at $(x_1,x_2) = (0,\ 60)$: $$\boxed{x_1 = 0\ \text{{t/yr A}},\quad x_2 = 60\ \text{{t/yr B}},\quad Z_{{\max}} = 240\ \text{{MU/yr}}.}$$ Product B is so light on both chemicals that Chemical 1 (60 t of B) is the only binding limit; making any A would displace the more profitable B.
(ii) Feasibility region. The region and the optimal vertex are shown below.

Figure 4 — Feasibility region for Q7 (shaded). The two chemical constraints cap the region from above; the labour line is the lower bound. Profit Z increases toward the upper right, so the optimum is the corner (0, 60).

Quantity	Result
Product A	$x_1 = 0$ t/yr
Product B	$x_2 = 60$ t/yr
Maximum profit	$Z_{{\max}} = 240$ MU/yr
Binding constraint	Chemical 1 ($3x_2 = 180$)

Concept & Theory

Topic. Two-variable linear programming solved graphically (corner-point / vertex theorem).

Key relations. An LP optimum, if it exists, occurs at a vertex of the feasible polytope; evaluate $Z$ at each vertex and pick the best. The $\ge$ labour constraint sets a lower boundary; the two $\le$ resource constraints set upper boundaries.

Why this works. A linear objective has straight, parallel contours, so its maximum over a convex polygon is pushed to a corner. Enumerating the (few) vertices is exact for two variables and needs no simplex tableau. The "make only B" result reflects B's very low resource intensity per unit profit.

Common pitfall. Treating the labour requirement as an upper limit ($\le$) instead of the stipulated minimum ($\ge$), which would wrongly delete the lower boundary and change the feasible set.

Source. Edgar, Himmelblau & Lasdon, Optimization of Chemical Processes, ch. 7 (linear programming); Rao, Engineering Optimization, ch. 3.

Question 7 of 7: Linear-Programming Product Mix for Maximum Profit

Question 7: Linear-Programming Product Mix for Maximum Profit (20 marks)