Math, asked by dhruvikathiriya, 3 months ago

A Factory Manufactures Two Products A And B. To Manufacture One Unit Of A, 1.5 Machine Hours And 2.5 Labour Hours Are Required. To Manufacture Product B, 2.5 Machine Hours And 1.5 Labour Hours Are Required. In A Month, 300 Machine Hours And 240 Labour Hours Are Available. Profit Per Unit For A Is Rs. 50 And For B Is Rs.40. Formulate As Lpp.​

Answers

Answered by pulakmath007
10

SOLUTION

TO DETERMINE

  • A Factory Manufactures Two Products A And B.

  • To Manufacture One Unit Of A, 1.5 Machine Hours And 2.5 Labour Hours Are Required.

  • To Manufacture Product B, 2.5 Machine Hours And 1.5 Labour Hours Are Required.

  • In A Month, 300 Machine Hours And 240 Labour Hours Are Available.

  • Profit Per Unit For A Is Rs. 50 And For B Is Rs.40.

Formulate As LPP

EVALUATION

Let the Factory manufactures x units of product A and y units of product B

 \sf{ \therefore \:  \:  \: x \geqslant 0 \:  \: and \:  \: y \geqslant 0}

To Manufacture One Unit Of A and one unit of product B 1.5 Machine Hours and 2.5 Machine Hours are Required respectively

Total Machine Hours required

= ( 1.5x + 2.5y ) hours

It is given that 300 Machine Hours are Available.

 \sf{ \therefore \:  \: 1.5x + 2.5y \leqslant 300}

 \sf{ \implies \:  \: 3x + 5y \leqslant 600}

Again To Manufacture One Unit Of A and one unit of product B 2.5 Labour Hours and 1.5 Labour Hours are Required respectively

Total Labour Hours required

= ( 2.5x + 1.5y ) hours

It is given that 240 Labour Hours are Available.

 \sf{ \therefore \:  \: 2.5x + 1.5y \leqslant 240}

 \sf{ \implies \:  \: 5x + 3y \leqslant 480}

Again Profit Per Unit For A is Rs. 50 And For B is Rs.40

So the objective function is

Maximize z = 50x + 40y

Hence Converting the given problem into a Linear programming problem we get

 \sf{Maximize  \:  \: z  =  50x  +  40y}

Subject to :

 \sf{  3x + 5y \leqslant 600}

 \sf{  5x + 3y \leqslant 480}

 \sf{ with \:  \: x \geqslant 0 \:  \: and \:  \: y \geqslant 0}

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Answered by saidahfjfd
0

Answer:

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