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  • A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs

A manufacturer produces two Models of bikes - Model X and Model Y. Model X takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hours available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models X and Y respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models X and Y are Rs 1000 and Rs 500, respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.

Answers (1)

Let us say that the number of bikes per week of model X and Y are x and y respectively.

Assuming that model X takes 6 man-hours.

So, time taken by x bikes of model X = 6x hours.

Assuming that model Y takes 10 man-hours.

So, time taken by y bikes of model X = 10y hours.

So, the total man-hour that is available per week = 450

So, 6x + 10y ≤ 450

3x + 5y ≤ 225

The handling and the marketing cost of model X and Y is Rs 2000 and Rs 1000 per unit respectively.

So, total handling and marketing cost of x units of model X and y units of model y is 2000x + 1000y

The maximum amount that is available for handling and marketing per week is Rs 80000.

So, 2000x + 1000y \leq 80000

\Rightarrow 2x + y ? 80

Profits per unit for Models X and Y are Rs 1000 and Rs 500, Respectively.

Let total profit = Z

So, Z = 1000x + 500y

Also, as units will be positive numbers so x, y ≥ 0

So, we have,

Z = 1000x + 500y

With constraints,

\\ \\ 3x + 5y $ \leq $ 225\\ \\ 2x + y $ \leq $ 80\\ \\ x, y $ \geq $ 0\\ \\

In order to maximize Z, that is subject to constraints.

We need to convert it into equation:

\\ 3x + 5y $ \leq $ 225\\ \\ $ \Rightarrow $ 3x + 5y = 225\\ \\ 2x + y $ \leq $ 80\\ \\ $ \Rightarrow $ 2x + y = 80\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\

The graph for the same is given below:

ABCD being the feasible region.

The Value of Z as well as the final answer is given below.

The Value of Z at corner points A,B,C and D :

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=1000 x+500 y \\ \hline A(0,45) & Z=0+45(500)=22500 \\ \hline B(25,30) & Z=25(1000)+30(500)=4000 \rightarrow \max \\ \hline C(40,0) & Z=40(1000)+0=40000 \rightarrow \max \\ \hline D(0,0) & Z=0+0=0 \\ \hline \end{array}

So, value of Z is maximum on-line BC, the maximum value is 40000 . So manufacturer must produce 25 number of models X and 30 number of model Y.

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