Question

(b) A company is involved in the production of two items (X and Y). The resources need to produce X and Y are twofold, namely machine time for automatic processing and craftsman time for hand finishing. The table below gives the number of minutes required for each item:

Machine time Craftsman time
Item X 13 20
Item Y 19 29

The company has 40 hours of machine time available in the next working week but only 35 hours of craftsman time. Machine time is costed at £10 per hour worked and craftsman time is costed at £2 per hour worked. Both machine and craftsman idle times incur no costs. The revenue received for each item produced (all production is sold) is £20 for X and £30 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer. Formulate the problem of deciding how much to produce per week as a linear program hence make the decision.

Answer 1

Answer on Question #56576 – Math – Other

A company is involved in the production of two items (X and Y). The resources need to produce X and Y are twofold, namely machine time for automatic processing and craftsman time for hand finishing. The table below gives the number of minutes required for each item:

The company has 40 hours of machine time available in the next working week but only 35 hours of craftsman time. Machine time is costed at £10 per hour worked and craftsman time is costed at £2 per hour worked. Both machine and craftsman idle times incur no costs. The revenue received for each item produced (all production is sold) is £20 for X and £30 for Y. The company has a specific contract to produce 10 items of X per week for a particular customer. Formulate the problem of deciding how much to produce per week as a linear program hence make the decision.

Solution

Let $x$ be the number of items of $X$, $y$ be the number of items of $Y$.

Then the LP is

maximise

$$20x + 30y - 10(\text{machine time worked}) - 2(\text{craftsman time worked})$$

subject to:

$$13x + 19y \leq 40(60) \text{ machine time}$$ $$20x + 29y \leq 35(60) \text{ craftsman time}$$ $$x \geq 10 \text{ contract}$$ $$x, y \geq 0$$

so that the objective function becomes

maximise

$$20x + 30y - \frac{10(13x + 19y)}{60} - \frac{2(20x + 29y)}{60}$$

i.e. maximise

$$17.1667x + 25.8667y$$

subject to:

$$13x + 19y \leq 2400$$ $$20x + 29y \leq 2100$$ $$x \geq 10$$

x,y\geq 0

It is plain from the diagram below that the maximum occurs at the intersection of $x = 10$ and

$20x + 29y\leq 2100$

Solving simultaneously, rather than by reading values off the graph, we have that $x = 10$ and $y = 65.52$ with the value of the objective function being £1866.5.

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