The Handy-Dandy Company makes three types of kitchen appliances ($A$, $B$ and $C$). To make each of these appliance types, just two inputs are required - labour and materials. Each unit of $A$ made requires $7$ hours of labour and $4$ kg of materials; for each unit of $B$ made the requirements are $3$ hours of labour and $4$ kg of materials, while for $C$ the unit requirements are $6$ hours of labour and $5$ kg of material. The company expects to make a profit of $40$€ for every unit of $A$ sold, while the profit per unit for $B$ and $C$ are $20$€ and $30$€ respectively. Given that the company has available to it $150$ hours of labour and $200$ kg of material each day, formulate this as a linear programming problem.
Step 1: Identify decision variables (what decisions can be made?)
For this production mix problem the company can decide on how many units of appliances $A$, $B$ and $C$ to make. Therefore, the decision variables are, say:
Let $x_1$ = number of units of appliance $A$ to make per day
Let $x_2$ = number of units of appliance $B$ to make per day
Let $x_3$ = number of units of appliance $C$ to make per day
Step 2: Identify restrictions or constraints
Evidently, $x_1 \geq 0, x_2 \geq 0, x_3 \geq 0$.
Also, there are limitations (or constraints) on labour and materials:
Labour (per day): $7x_1 + 3x_2 + 6x_3 \leq 150$
Materials (per day): $4x_1 + 4x_2 + 5x_3 \leq 200$
Note: Here we have assumed linearity (scaling and additivity).
Step 3: Identify “best” criterion
It seems clear that profit should be maximised, that is the objective function is to maximise $40x_1 + 20x_2 + 30x_3$.
Note: Here also we have assumed linearity (scaling and additivity).
I'm having trouble understanding how they can up with the Maximise $40x_1 + 20x_2 + 30x_3$ as the final answer. Can someone explain how they go this. Thank You.