Question 9.2: A company manufactures two types of wrought iron gates. The ......

(a) For x type I gates and y type II gates, the number of labour-hours, revenue and profit are calculated by multiplying the requirements for one gate of each type, as given in Table 9.3, by x and y, respectively. This is given in Table 9.4.

The constraints are
Welding                         6x + 2y ≤ 840                        (9.5)
Finishing                       2x + y ≤ 300                          (9.6)
Administration             x + y ≤ 250 x ≥ 0, y ≥ 0       (9.7)
The equations for total revenue and profit are
Total revenue:              TR = 120x + 95y                   (9.8)
Profit:                             π = 55x + 25y                        (9.9)

(b) Plot the constraints, by calculating the points of intersection with the axes.

These lines are sketched in Figure 9.4. Since the inequalities are all less than or equal to a limit, the feasible region is the overlap region of all three constraints, (c) TR = 120x + 95y.

To plot one revenue line, choose any value of TR, for example, TR = 9500. This line (called an isorevenue line) cuts the y-axis at y = 100.00 (when x is zero) and the x-axis at x = 79.167 (when y = 0). See Figure 9.5. As revenue increases, the isorevenue lines move upwards away from the origin.
Profit: π = 55x + 55y.
To plot one isoprofit line, choose any value of π, for example, π = 1100. This line cuts the y-axis at y = 44 (when x is zero) and the x-axis at x = 20 (when y = 0). As profit increases, the isoprofit lines move upwards away from the origin. See Figure 9.5.

(d) For maximum revenue, it is required that the isorevenue line be as ‘high’ as possible, but contain one point in the feasible region. From Figure 9.6(a), this point is at B, where x = 50, y = 200. For maximum profit, it is required that the isoprofit line be as ‘high’ as possible, but still contain one point in the feasible region. From Figure 9.6(b), this point is at C, where x = 120, y = 60.

The answer is confirmed algebraically by solving the appropriate pairs of equations at points B and C.
Point B: Maximum revenue is at the intersection of equations (2), finishing, and (3), administration.

(2) 2x + y = 300
(3) x + y = 250      The solution of these equations is x = 50, y = 200

Point C: The maximum profit is at the intersection of equations (1), welding, and (2), finishing.

(1) 6x + 2y = 840
(2) 2x + y = 300     The solution of these equations is x = 120, y = 60

To demonstrate the extreme point theorem, calculate profit and revenue at all corner points (just to check the answers above).

(e) When revenue is maximised (x = 50, y = 200), there are 140 unused welding hours:

When profit is maximised (x = 120, y = 60) there are 70 unused administration hours: