Question

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 timeItem X 13 20 Y 19 29The 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.​

Answer 1

SOLUTION

TO DETERMINE

Formulate the problem of deciding how much to produce per week as a linear program.

EVALUATION

Let number of item X = x

Number of item Y = y

The company has a specific contract to produce 10 items of X per week for a particular customer

$\displaystyle\sf{ \therefore \: x \geqslant 10} \: \: \: - - - (1)$

Work time for machine

$\displaystyle\sf{ = (13x + 19y) \: \: minutes}$

$\displaystyle\sf{ = \frac{13x + 19y}{60} \: \: hours}$

$\displaystyle\sf{ \therefore \: \: \frac{13x + 19y}{60} \leqslant 40}$

$\displaystyle\sf{ \implies \: \: 13x + 19y \leqslant 2400 \: \: - - - (2)}$

Work time for Craftsman

$\displaystyle\sf{ = (20x + 29y) \: \: minutes}$

$\displaystyle\sf{ = \frac{20x + 29y}{60} \: \: hours}$

$\displaystyle\sf{ \therefore \: \: \frac{20x + 29y}{60} \leqslant 35}$

$\displaystyle\sf{ \implies \: \: 20x + 29y \leqslant 2100} \: \: \: - - - (3)$

Let z be the profit

Then

z = ( 20x + 30y ) - 10 × Work time for machine - 2 × Work time for Craftsman

$\displaystyle\sf{ \therefore \: z =(20x + 30y) - 10 \times \frac{13x + 19y}{60} - 2 \times \frac{20x + 29y}{60}}$

$\displaystyle\sf{ \therefore \: z = \frac{1030x + 1552y}{60} }$

Thus converting the given problem into linear programming problem we get

$\displaystyle\sf{ Maximize \: \: z = \frac{1030x + 1552y}{60} }$

Subject to :

$\displaystyle\sf{ \: x \geqslant 10} \: \: \:$

$\displaystyle\sf{ \: 13x + 19y \leqslant 2400 \: \: }$

$\displaystyle\sf{ \: 20x + 29y \leqslant 2100} \: \: \:$

$\sf{With \: \: x \geqslant 0 \: \: and \: \: y \geqslant 0}$

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