Question 19.SP.1: A small construction firm specializes in building and sellin...
A small construction firm specializes in building and selling single-family homes. The firm offers two basic types of houses, model A and model B. Model A houses require 4,000 labor hours, 2 tons of stone, and 2,000 board feet of lumber. Model B houses require 10,000 labor hours, 3 tons of stone, and 2,000 board feet of lumber. Due to long lead times for ordering supplies and the scarcity of skilled and semiskilled workers in the area, the firm will be forced to rely on its present resources for the upcoming building season. It has 400,000 hours of labor, 150 tons of stone, and 200,000 board feet of lumber. What mix of model A and B houses should the firm construct if model A yields a profit of $3,000 per unit and model B yields $6,000 per unit? Assume that the firm will be able to sell all the units it builds.
1. Formulate the objective function and constraints:1
Maximize Z = 3,000A + 6,000B
Subject to
Labor 4,000A+ 10,000B ≤ 400,000 labor hours
Stone 2A+ 3B ≤ 150 tons
Lumber 2,000A+ 2,000B ≤ 200,000 board feet
A, B ≤ 0
2. Graph the constraints and objective function, and identify the optimum corner point (see graph). Note that the lumber constraint is redundant: It does not form a boundary of the feasible solution space.
3. Determine the optimal quantities of models A and B, and compute the resulting profit. Because the optimum point is at the intersection of the stone and labor constraints, solve those two equations for their common point:
\begin{array}{r c}\begin{matrix} Labor 4,000A + 10,000B \\ − 2,000 × (Stone 2A + 3B\end{matrix} & \begin{matrix} = 400,000 \\ = 150)\end{matrix} \\ \hline \begin{matrix} 4,000B\end{matrix}&\begin{matrix}= 100,000\end{matrix} \end{array}B = 25
Substitute B = 25 in one of the equations, and solve for A:
2A + 3(25) = 150 A = 37.5
Z = 3,000(37.5) + 6,000(25) = 262,500
4. We could have used the enumeration approach to find the optimal corner point. The corner points and the value of the objective function at each corner point are:
A = 0, B = 40 (found by inspection); Z = 3,000(0) + 6,000(40) = 240,000
A = 37.5, B = 25 (found using simultaneous equations); Z = 262,500 (see step 3)
A = 75, B = 0 (found by inspection); Z = 3,000(75) + 6,000(0) = 225,000
The best value of Z is 262,500 (because this is a maximization problem), so that indicates that the optimal corner point is A = 37.5, B = 25.
1For the sake of consistency, we will assign to the horizontal axis the first decision variable mentioned in the problem. In this case, variable A will be represented on the horizontal axis and variable B on the vertical axis.