(Diet problem) A diet is to contain atleast 80 units of vitamin A and 100 units of minerals. Two foods $F_1$ and $F_2$ are available. Food $F_1$ costs ₹4 per kg and $F_2$ costs ₹5 per kg. One kg of food $F_1$ contains 3 units of vitamin A and 4 units of minerals. One kg of food $F_2$ contains 6 units of vitamin A and 3 units of minerals. We wish to find the minimum cost for diet that consists of mixture of these two foods and also meets the minimum nutritional requirements. Formulate this as a linear programming problem.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Minimize $Z=4x+5y$, Subject to: $3x+6y≥80,4x+3y≥100,x,y≥0$
Let the mixture consist of x kg of food $F_1$ and y kg of food $F_2$.
We make the following nutritional requirement table from given data:
|
Resources |
Food (in kg) |
Requirement (in units) |
|
|
$F_1$ |
$F_2$ |
||
|
Vitamin A (units/kg) |
3 |
6 |
80 |
|
Vitamin B (units/kg) |
4 |
3 |
100 |
Minimum requirement of Vitamin A is 80 units,
therefore, $3x+6y≥80$
Similarly, the minimum requirement of minerals is 100 units,
therefore, $4x+3y≥100$
Also, $x≥0,y≥0$
As cost of food $F_1$ is ₹4 per kg, and the cost of food $F_2$ is ₹5 per kg, the total cost of purchasing x kg of food $F_1$ and y kg of food $F_2$ is $Z = 4x+5y$, which is the objective function.
Hence, the mathematical formulation of the L.P.P. is:
Minimize $Z = 4x + 5y$ subject to the constraints
$3x + 6y ≥ 80, 4x + 3y ≥ 100, x ≥ 0, y ≥ 0$.