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

A company manufactures two types of wrought iron gates. The number of labour-hours required to produce each type of gate, along with the maximum number of hours available, are given in Table 9.3.

(a) Write down expressions for:

(i) The constraints.
(ii) Total revenue.
(iii) Profit.

(b) Plot the constraints and shade in the feasible region.
(c) Plot the revenue and profit lines, indicating the direction of increasing revenue and profit.
(d) Determine, graphically, the number of gates which should be produced and sold to maximise:

(i) Revenue.
(ii) Profit.
Confirm these answers algebraically, using the extreme point theorem.

(e) Calculate the number of labour-hours which are not used when:

(i) Revenue is maximised.
(ii) Profit is maximised.

 Table 9.3 Requirements for gates type I and II Welding Finishing Admin. Selling price (£) Profit (£) Type I gate 6 2 1 120 55 Type II gate 2 1 1 95 25 Max. hours available 840 300 250
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(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:

 Table 9.4 Requirements for x type I gates and y type II gates Welding Finishing Admin. Selling price (£) Profit (£) x type I gates 6 x 2 x x 120x 55x y type II gates 2 y 1 y y 95y 25y Max. hours available 840 300 250

 Activity Equation Cuts x-axis when y =0 Cutsy-axis when x = 0 Welding (1) 6x + 2y = 840 6x = 840 → x = 140 2y = 840 → y = 420 Finishing (2) 2x + y = 300 2x = 300 → x = 150 y = 300 Administration (3) x + y = 250 x = 250 y = 250 x ≥ 0, y ≥ 0 First quadrant

 Point (x, y) TR = 120x + 95y Profit = 55x + 25y A (x = 0, y = 250) TR = 120(0) + 95(250) = 23 750 Profit = 55(0) + 25(250) = 6250 B (x = 50, y = 200) TR = 120(50) + 95(200) = 25 000* Profit = 55(50) + 25(200) = 7750 C (x = 120, y = 60) TR = 120(120) + 95(60) = 20 100 Profit = 55(120) + 25(60) = 8100* D (x = 140, y = 0) TR = 120(140) + 95(0) = 16 800 Profit = 55(140) + 25(0) = 7700 *Maximum value, subject to constraints

 Welding Finishing Admin. 50 type I gates 6 (50) 2 (50) 1 (50) 200 type II gates 2 (200) 1 (200) 1 (200) Hours used 700 300 250 Max. hours available 840 300 250 Unused hours 140 0 0

 Welding Finishing Admin. 120 type I gates 6 (120) 2 (120) 1(120) 60 type II gates 2 (60) 1 (60) 1(60) Hours used 840 300 180 Max. hours available 840 300 250 Unused hours 0 0 70

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