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Using Autograph in the Classroom Linear Programming Linear Programming Linear Programming is a technique used to solve problems which involve maximising or minimising an objective function subject to constraints in the form of inequalities. With two variables (x and y) a graphical approach is a good way to solve the problem, and is easily achieved using a graph plotter such as Autograph. Example The problem A factory produces two types of drink, an ‘energy’ drink and a ‘refresher’ drink. The day’s output is to be planned. Each drink requires syrup, vitamin supplement and concentrated flavouring, as shown in the table. Vitamin Concentrated Syrup supplement flavouring 1 litre of energy drink 0.25 litres 0.4 units 6 cc 1 litre of refresher drink 0.25 litres 0.2 units 4 cc Availabilities 250 litres 300 units 4.8 litres The last row in the table shows how much of each ingredient is available for the day’s production. Energy drink sells at £1 per litre : Refresher drink sells at 80 p per litre How can the factory manager decide how much of each drink to make? The formulation Let x represent number of litres of energy drink Let y represent number of litres of refresher drink Objective function: Maximise x + 0.8y Subject to: Syrup constraint: 0.25x + 0.25y 250 x + y 1000 Vitamin supplement constraint: 0.4x + 0.2y 300 2x + y 1500 Concentrated flavouring constraint: 6x + 4y 4800 3x + 2y 2400 Bob Francis 1 March 2006 Using Autograph in the Classroom Linear Programming The solution Draw the graphs representing the constraints; shade out the regions that do NOT satisfy the inequalities. 1600 y 1400 1200 A 1000 800 B 600 400 C 200 D x - 200 O 200 400 600 800 1000 1200 - 200 The unshaded polygon is called the feasible region. Now work out the value of the objective function x + 0.8y at each of the vertices of the feasible region: At O: x + 0.8y = 0 + 0.8 0 = 0 At A: x + 0.8y = 0 + 0.8 1000 = 800 At B: x + 0.8y = 400 + 0.8 600 = 880 At C: x + 0.8y = 600 + 0.8 300 = 840 At D: x + 0.8y = 750 + 0.8 0 = 750 The objective function, x + 0.8y, is a maximum at B(400, 600), giving an income of £880. Alternatively, graph the objective function, x + 0.8y = k, and see how large you can make the value of k without leaving the feasible region – this gives a value of k = 880 at point B. Bob Francis 2 March 2006 Using Autograph in the Classroom Linear Programming Problem for you to solve 1. A factory produces two types of toy: bicycles and trucks. In the manufacturing process of these toys, three machines are used. These are a moulder, a lathe and an assembler. The table shows the length of time (in hours) needed for each toy. Moulder Lathe Assembler Bicycle 1 3 1 Truck 0 1 1 The moulder can be operated for 3 hours per day, the lathe for 12 hours per day, and the assembler for 7 hours per day. Each bicycle made gives a profit of £15 and each truck made gives a profit of £11. Let x be the number of bicycles made per day and y be the number of trucks made per day. (i) Write down the objective function to be maximised and three constraints. (ii) Use Autograph to graph the inequalities, display the feasible region and find the combination of x and y that maximises the profit. Bob Francis 3 March 2006 Using Autograph in the Classroom Linear Programming Problem for you to solve 2. An oil company has two refineries. The amount of oil, in barrels produced per day, is given in the table. High grade Medium grade Low grade Refinery 1 100 200 300 Refinery 2 200 100 200 An order is received for 2000 barrels of high grade oil, 2000 barrels of medium grade oil and 3600 barrels of low grade oil. Refinery 1 costs £10000 per day to operate and Refinery 2 costs £9000 per day to operate. Let x be the number of days for which refinery 1 is operated and y be the number of days for which refinery 2 is operated. (i) Write down the objective function to be minimised and three constraints. (ii) Use Autograph to graph the inequalities, display the feasible region and find the combination of x and y that minimises cost. Bob Francis 4 March 2006

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posted: | 10/31/2011 |

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