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.