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# A firm manufactures three products A,B and C. Time to manufacture product A is twice that for B and thrice that for C and if the entire labour is engaged in making product A,1600 units of this product can be produced. These products are to be produced in the ratio 3:4:5. There is demand for at least 300,250 and 200 units of products A,B and C and the profit earned per unit is Rs.90, Rs40 and Rs.30 respectively.Raw materialRequirement per unit product(Kg) ARequirement per unit product(Kg) BRequirement per unit product(Kg) CTotal availability (kg)P6525,000Q4736,000Formulate 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 ${x_1},{x_2},{x_3} \geqslant 0$  Verified
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Hint:To get the formulation of linear programming problem, we will collect the relevant information from the question with the given constraints. Then we will write them systematically with appropriate variables to get the LPP equations.

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 ${x_1},{x_2},{x_3} \geqslant 0$

So, the correct answer is “Option B”.

Note:Formulation of linear programming problem, which is the part of optimization problem, careful collection of the facts and hence conversion into constraints is very important. Then only their solutions can be obtained correctly.