Login/Sign Up
Home » A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, and , are available at a cost of and per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of \pm 4 and \pm 3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.
CBSE | Class 12 | Exercise 33B | Linear Programming | Maths | RS Aggarwal
A diet for a sick person must contain at least 4000 units of vitamins, 50 units of mineral and 1400 calories. Two food, A and B, are available at a cost of \pm 4 and \pm 3 per unit respectively. If one unit of A contains 200 units of vitamins, 1 unit of mineral and 40 calories, and 1 unit of B contains 100 units of vitamins, 2 units of mineral and 40 calories, find what combination of foods should be used to have the least cost.

Let x and y be number of units of food A and B.
\therefore According to the question,
200 x+100 y \geq 4000, x+2 y \geq 50,40 x+40 y \geq 1400, x \geq 0, y \geq 0
Minimize Z=4 x+3 y
The feasible region determined 200 x+100 y \geq 4000, x+2 y \geq 50,40 x+40 y \geq 1400, x \geq 0, y \geq 0 is given by

The feasible region is unbounded. The corner points of feasible region are A(0,40), B(5,30), C(20,15), D(50,0). The value of Z at corner points are

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $Z=4 x+3 y$ & \\ \hline $\mathrm{A}(0,40)$ & 120 & \\ \hline $\mathrm{B}(5,30)$ & 110 & Minimum \\ \hline $\mathrm{C}(20,15)$ & 125 & \\ \hline $\mathrm{D}(50,0)$ & 200 & \\ \hline \end{tabular}\]

The minimum value of Z is 110 at point (5,30).
Hence, the diet should contain 5 units of food A and 30 units of food B for the least cost.

i

More Material