Question

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 costs Rs.4. How many packets of mixed from S and T should the company purchase to honour the contract requirement and yet minimize cost? Make a LPP and solve graphically

Answer 1

Let x and y units of packet of mixes are purchased from S and T respectively. If z is total cost then
z = 10x + 4y ...(i)
isobjective function which we have to minimize
Here constraints are.

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

For coordinate of P

Point of intersection of
2x + y = 60 ...(vi)
and 4x + y = 80 ...(vii)
(vi) – (vii) => 2x + y - 4x - y = 60 - 80
=> -2x = -20 Þ x = 10
=>y = 40
co-ordinate 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 ...(viii)
Since the graph of inequality (viii) 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.