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A company manufactures two types of sweaters type and B. It costs $₹ 360$ to make a type A sweater and $₹ 120 \mathrm{t}$ make a type B sweater. The company can make atmost 30 sweaters and spend atmost $₹ 72000$ a day. The number 0 sweaters of type B cannot exceed the number of sweater of type A by more than 100 . The company makes a profit o $₹ 200$ for each sweater of type A and B ₹ 120 for ever: sweater of type B.
Formulate this problem as a LPP to maximise the profit to the company
Formulate this problem as a LPP to maximise the profit to the company
Solution:
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Verified Answer
Let the company manufactures $\mathrm{x}$ number of type A sweaters and y number of type $B$ sweaters.
According to the question
$$
\begin{aligned}
&360 x+120 y \leq 72000 \\
&\Rightarrow 3 x+y \leq 600 \\
&x+y \leq 300 \\
&x+100 \geq y \\
&\Rightarrow x-y \geq-100 \\
&x \geq 0, y \geq 0
\end{aligned}
$$
And the objective function to be maximised $\operatorname{Profit}(Z)=200 x+120 y$
Thus, the required LPP to maximise the profit is Maximise $Z=200 x+120 y$ is subject to constraints.
$$
\begin{aligned}
&3 x+y \leq 600 \\
&x+y \leq 300 \\
&x-y \geq-100 \\
&x \geq 0, y \geq 0
\end{aligned}
$$
According to the question
$$
\begin{aligned}
&360 x+120 y \leq 72000 \\
&\Rightarrow 3 x+y \leq 600 \\
&x+y \leq 300 \\
&x+100 \geq y \\
&\Rightarrow x-y \geq-100 \\
&x \geq 0, y \geq 0
\end{aligned}
$$
And the objective function to be maximised $\operatorname{Profit}(Z)=200 x+120 y$
Thus, the required LPP to maximise the profit is Maximise $Z=200 x+120 y$ is subject to constraints.
$$
\begin{aligned}
&3 x+y \leq 600 \\
&x+y \leq 300 \\
&x-y \geq-100 \\
&x \geq 0, y \geq 0
\end{aligned}
$$
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