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A manufacturer produces two models of bikes-model $\mathrm{X}$ and model $Y$. Model $\mathrm{X}$ takes a 6 man-hours to make per unit, while model $Y$ takes 10 man hours per unit. There is a total of 450 man-hour available per week. Handling and marketing costs are $₹ 2000$ and $₹ 1000$ per unit for models $\mathrm{X}$ and $\mathrm{Y}$, respectively. The total funds available for these purposes are $₹ 80000$ per week. Profits per unit for models $\mathrm{X}$ and $\mathrm{Y}$ are $₹ 1000$ and $₹ 500$, respectively. How many bikes of each model should the manufacturer produce, so as to yield a maximum profit? Find the maximum profit.
Solution:
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Verified Answer
Let $\mathrm{x}$ be number of models $\mathrm{X}$ and $\mathrm{y}$ be number of models $\mathrm{Y}$ bikes. Then according to question
$$
\begin{aligned}
\therefore & 6 \mathrm{x}+10 \mathrm{y} \leq 450 \\
\Rightarrow & 3 \mathrm{x}+5 \mathrm{y} \leq 225 \\
& 2000 \mathrm{x}+1000 \mathrm{y} \leq 80000 \\
\Rightarrow & 2 \mathrm{x}+\mathrm{y} \leq 80
\end{aligned}
$$
Also, $x \geq 0, y \geq 0$
$\therefore$ Required LPP is
Maximise $\mathrm{Z}=1000 \mathrm{x}+500 \mathrm{y}$
Subject to $3 x+5 y \leq 225,2 x+y \leq 80, x \geq 0, y \geq 0$
$(40,0),(25,30)$ and $(0,45)$
Solving $3 x+5 y=225$ and $2 x+y=80$, we get $x=25, y=30$
$$
\begin{array}{|l|l|}
\hline \text { Corner points } & \mathbf{Z}=\mathbf{1 0 0 0 x}+\mathbf{5 0 0 y} \\
\hline(0,0) & 0 \\
(40,0) & 40000 \leftarrow \text { Maximum } \\
(25,30) & 25000+15000 \\
& =40000 \leftarrow \text { Maximum } \\
(0,45) & 22500 \\
\hline
\end{array}
$$
Since, it is asked that each model bikes should be produced, the solution of the LPP is $x=25$ and $y=30$.
$$
\begin{aligned}
\therefore & 6 \mathrm{x}+10 \mathrm{y} \leq 450 \\
\Rightarrow & 3 \mathrm{x}+5 \mathrm{y} \leq 225 \\
& 2000 \mathrm{x}+1000 \mathrm{y} \leq 80000 \\
\Rightarrow & 2 \mathrm{x}+\mathrm{y} \leq 80
\end{aligned}
$$
Also, $x \geq 0, y \geq 0$
$\therefore$ Required LPP is
Maximise $\mathrm{Z}=1000 \mathrm{x}+500 \mathrm{y}$
Subject to $3 x+5 y \leq 225,2 x+y \leq 80, x \geq 0, y \geq 0$
$(40,0),(25,30)$ and $(0,45)$
Solving $3 x+5 y=225$ and $2 x+y=80$, we get $x=25, y=30$
$$
\begin{array}{|l|l|}
\hline \text { Corner points } & \mathbf{Z}=\mathbf{1 0 0 0 x}+\mathbf{5 0 0 y} \\
\hline(0,0) & 0 \\
(40,0) & 40000 \leftarrow \text { Maximum } \\
(25,30) & 25000+15000 \\
& =40000 \leftarrow \text { Maximum } \\
(0,45) & 22500 \\
\hline
\end{array}
$$
Since, it is asked that each model bikes should be produced, the solution of the LPP is $x=25$ and $y=30$.
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