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  • A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B can

A company manufactures two types of sweaters: type A and type B. It costs Rs 360 to make a type A sweater and Rs 120 to make a type B sweater. The company can make at most 300 sweaters and spend at most Rs 72000 a day. The number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100. The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B. Formulate this problem as a LPP to maximise the profit to the company.

Answers (1)

Let’s take an assumption that the company manufactures x number of type A sweaters and y number of type B. We have made a table for your understanding which is given below:

\begin{array}{|l|l|l|l|} \hline & \begin{array}{l} \text { Type A Sweaters } \\ (\mathrm{X}) \end{array} & \begin{array}{l} \text { Type B Sweaters } \\ (\mathrm{Y}) \end{array} & \\ \hline \text { cost per day } & \text { Rs. } 360 & \text { Rs. } 120 & \text { Rs. } 72000 \\ \hline \begin{array}{l} \text { Number of } \\ \text { Sweaters } \end{array} & 1 & 1 & 300 \\ \hline \text { Profit } & \text { Rs.200 } & \text { Rs.120 } & \\ \hline \end{array}

If we look at the table, we can see that the profit becomes Z=200x+120y

If we have to maximize the profit, then maximize Z=200x+120y

The constraints so obtained, i.e., subject to the constraints,
The company spends at most Rs 72000 a day.
\therefore 360 x+120 y \leq 72000
Divide throughout by 120, we get

\Rightarrow 3 x+y \leq 600 \ldots...(i)

Also, company can make at most 300 sweaters.:

x+y \leq 300 \ldots (ii)

Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100
i.e., y-x \leq 100

\Rightarrow y \leq 100+x \ldots \ldots \ldots .....(iii)
And x \geq 0,y\geq 0 [non-negative constraint]
So, to maximize profit we have to maximizeZ=200x+120y subject to
\\ 3 x+y \leq 600 \\$$ \begin{array}{l} x+y \leq 300 \\ y \leq 100+x \\ x \geq 0, y \geq 0 \end{array}

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