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Raw material | Requirement per unit product(Kg) A | Requirement per unit product(Kg) B | Requirement per unit product(Kg) C | Total availability (kg) |

P | 6 | 5 | 2 | 5,000 |

Q | 4 | 7 | 3 | 6,000 |

Formulate the problem as a linear programming problem and find all the constraints for the above product mix problem.

A. \[3{x_1} - 4{x_2} = 0\] and \[5{x_2} - 4{x_3} = 0\] where \[{x_1},{x_2},{x_3} \geqslant 0\]

B. $4{x_1} - 3{x_2} = 0$ and $5{x_2} - 4{x_3} = 0$ where ${x_1},{x_2},{x_3} \geqslant 0$

C. $4{x_1} - 3{x_2} = 0$ and $4{x_2} - 5{x_3} = 0$ where ${x_1},{x_2},{x_3} \geqslant 0$

D. $4{x_1} - 3{x_2} = 0$ and $5{x_2} - 4{x_3} = 0$ where ${x_1},{x_2},{x_3} \geqslant 0$

Answer

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Let ${x_1},{x_2},{x_3}$are the number of units of products A,B and C to be manufactured .

Thus, the objective is to maximize the profit.

Mathematically, maximize \[Z = 90{x_1} + 40{x_2} + 30{x_3}\]

We can formulate the constraints as follows:

For the raw material P,\[6{x_1} + 5{x_2} + 2{x_3} \leqslant 5000\]

For raw material Q,\[4{x_1} + 7{x_2} + 3{x_3} \leqslant 6000\]

Product B requires $\dfrac{1}{2}$ and product C requires$\dfrac{1}{3}$the time required for product A.

Now, $\dfrac{t}{2}$ and$\dfrac{t}{3}$ will be the times in hours to produce B and C and since for 1600 units of A we need time 1600t hours. So, its constraint will be,

\[

t{x_1} + \dfrac{t}{2}{x_2} + \dfrac{t}{3}t{x_3} \leqslant 1600t \\

\Rightarrow {x_1} + \dfrac{{{x_2}}}{2} + \dfrac{{{x_3}}}{3} \leqslant 1600 \\

\Rightarrow 6{x_1} + 3{x_2} + 2{x_3} \leqslant 9600 \\

\]

Market demand will require that,

\[{x_1} \geqslant 300,{x_2} \geqslant 250,\;and\;{x_3} \geqslant 200\;\]

Here, products A,B and C should be in the ratio 3:4:5,

So, ${x_1}:{x_2}:{x_3} = 3:4:5$

$

\Rightarrow \dfrac{{{x_1}}}{3} = \dfrac{{{x_2}}}{4} \\

\\

$

And \[

\dfrac{{{x_2}}}{4} = \dfrac{{{x_3}}}{5} \\

\\

\]

$\therefore $ These are the following constraints finally,

$4{x_1} - 3{x_2} = 0$ and $5{x_2} - 4{x_3} = 0$ where ${x_1},{x_2},{x_3} \geqslant 0$