This is easiest to explain in an example:
A recycling plant produces 2 quanlities of paper, P1 and P2, using a mixture of scrap paper and pulverised wood. Each tonne of P1 paper requires 2 tonnes of scrap paper and 2 tonnes of wood. Each tonne of P2 requires 1 tonne of scrap paper and 3 tonnes of wood. Each day, there is a minimum of 16 tonnes of scrap paper and 24 tonnes of wood available. How can the plant maximise the total output of paper per day?
Stage 1: Define Variables
Let 'x' be the amount of P1 produced per day, and let 'y' be the amount of P2 produced each day.
Stage 2: Write down constraints
16 tonnes of scrap paper:
Therefore, we can write this expression: 2x + y ≤ 16
24 tonnes of wood:
Therefore, we can write this expression: 2x + 3y ≤ 24
Because this is a 'real life problem' we also know two further contraints, which must also be stated. These are, x ≥ 0 and y ≥ 0
Stage 3: Define the objective function [the quantity you are trying to maximise or minimise]
To maximise: T = x + y [you must say if you are maximising or minimising]
Stage 4: Draw the graph
Plot:
Plot these points using x = 0 and y = 0, as this is the quickest way. Then label the graphs
Shade in the areas which do not satisfy the equations so you can therefore identify the feasible region, that is, the unshaded area. [See Fig. 3]
Stage 5: Consider the objective function
Plot the line graph corresponding to one value of the objective function. Slide ruler to identify the vertex at which the objective funtion will have maximum value.
The maximum value occurs at B. B satisfies:
Solve the simultaneous equations: y = 4, x = 6. These are the co-ordinates of B.
So the maximum value occurs when x = 6 and y = 4. Therefore, T = 6 + 4 = 10
Note that if the objective function is parallel to the optimum solution line, there is no single solution. You must therefore say the line equation.