A company producing soft drinks has a contrast which requires a minimum of 80 units of chemical A and 60 , units of chemical B to go in each bottle of the drink. The chemical are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs Rs.10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs \mp4. How many mixes from X and Y should the company purchase to honor the contract requirement and yet minimize the cost?
A company producing soft drinks has a contrast which requires a minimum of 80 units of chemical A and 60 , units of chemical B to go in each bottle of the drink. The chemical are available in a prepared mix from two different suppliers. Supplier X has a mix of 4 units of A and 2 units of B that costs Rs.10, and the supplier Y has a mix of 1 unit of A and 1 unit of B that costs \mp4. How many mixes from X and Y should the company purchase to honor the contract requirement and yet minimize the cost?

Let x and y be number of mixes from suppliers X and Y.
\therefore According to the question,
4 x+y \geq 80,2 x+y \geq 60, x \geq 0, y \geq 0
Minimize Z=10 x+4 y
The feasible region determined by 4 x+y \geq 80,2 x+y \geq 60, x \geq 0, y \geq 0 is given by

The feasible region is unbounded .The corner points of feasible region are A(0,80), B(10,40), C(30,0).
The value of Z at corner points are

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $\mathrm{Z}=10 \mathrm{x}+4 \mathrm{y}$ & \\ \hline $\mathrm{A}(0,80)$ & 320 & \\ \hline $\mathrm{B}(10,40)$ & 260 & Minimum \\ \hline $\mathrm{C}(30,0)$ & 300 & \\ \hline \end{tabular}\]

The minimum value of Z is 260 at point (10,40).
Hence, the company should buy 10 mixes from supplier \mathrm{X} and 40 mixes from supplier \mathrm{Y} to minimize the cost.