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in Linear Programming by (15.3k points)

A company produces soft drinks that has a contract which requires that a minimum of 80 units of the chemical A and 60 units of the chemical B go into each bottle of the drink. The chemicals are available in prepared mix packets from two different suppliers. Supplier S had a packet of mix of 4 units of A and 2 units of B that costs Rs 10. The supplier T has a packet of mix of 1 unit of A and 1 unit of B that costs Rs 4. How many mix packets mix from S and T should the company purchase to honour the contract requirement and yet minimize cost? Make a LPP and solve graphically.

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1 Answer

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by (12.4k points)

Let x and y units of packet of mixes be purchased from S and T respectively. If Z is total cost then

Z = 10x + 4y ...(i)

is objective function, which we have to minimize.

Here, constraints are:

4x + y ≥ 80 ...(ii)

2x + y ≥ 60 ...(iii)

Also, x, y ≥ 0 ...(iv)

On plotting graph of above constraints or inequalities (ii), (iii) and (iv), we get shaded region having corner point A, P, B as feasible region.

For coordinate of P.

Point of intersection of

2x + y = 60 ...(v)

and 4x + y = 80 ...(vi)

(v) – (vi)

⇒ 2x + y – 4x – y = 60 – 80

⇒ –2x = –20

⇒ x = 10

⇒ y = 40

coordinate of P ≡ (10, 40)

Now the value of Z is evaluated at corner point in the following table

Since, feasible region is unbounded. Therefore we have to draw the graph of the inequality.

10x + 4y < 260 ...(vii)

Since, the graph of inequality (vii) does not have any point common.

So, the minimum value of Z is 260 at (10, 40).

i.e., minimum cost of each bottle is Rs 260 if the company purchases 10 packets of mixes from S and 40 packets of mixes from supplier T.

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