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) is objective function which we have to minimize. Here constraints are.
4x + y ≥ 80 ...(ii)
2x + y ≥ 60 ...(iii)
Also, x ≥ 0 ...(iv)
y ≥ 0 ...(v)
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 ⇒ x = 10
⇒ y = 40
Since co-ordinate of P = (10, 40)
Now the value of Z is evaluated at corner point in the following table
Corner points |
Z = 10x + 4y |
A(30, 0) |
300 |
P(10, 40) |
260 Minimum |
B(0, 80) |
320 |
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.