A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 of calories. Two foods *A* and *B*, are available at a cost of Rs 4 and Rs 3 per unit respectively. If one unit of *A* contains 200 units of vitamin, 1 unit of mineral and 40 calories and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 calories, find what combination of foods should be used to have the least cost?

#### Solution

Let the sick person takes *x* units and *y* units of food I and II respectively that were taken in the diet.

Since, per unit of food I costs Rs 4 and that of food II costs Rs 3.

Therefore, *x* units of food I costs Rs 4*x** *and *y* units of food II costs Rs 3*y*.

Total cost = Rs (4*x** *+ 3*y*)

Let Z denote the total cost

Then, Z = 4*x *+ 3*y*

If one unit of *A* contains 200 units of vitamin and one unit of food B contains 100 units of vitamin.

Thus, *x* units of food I and *y* units of food II contains 200*x* + 100*y* units of vitamin.

But a diet for a sick person must contain at least 4000 units of vitamins.

∴ \[200x + 100y \geq 4000\]

If one unit of *A* contains 1 unit of mineral and one unit of food B contains 2 units of mineral.

Thus, *x* units of food I and *y* units of food II contains* x* + 2*y* units of mineral.

But a diet for a sick person must contain at least 50 units of vitamins.

∴ \[x + 2y \geq 50\] If one unit of *A* contains 40 calories and one unit of food B contains 40 calories.

Thus, *x* units of food I and *y* units of food II contains* *40*x* + 40*y* units of calories.

But a diet for a sick person must contain at least 1400 calories.

So

Min

*Z*= 4

*x*

*+ 3*

*y*

subject to

\[x + 2y \geq 50\]

\[40x + 40y \geq 1400\]

\[x, y \geq 0\]

First, we will convert the given inequations into equations, we obtain the following equations:

200*x** *+ 100*y* = 4000, *x** *+2*y* = 50,* *40*x** *+ 40*y* =1400, *x* = 0 and* y* = 0

Region represented by 200*x** *+ 100*y* ≥ 4000:

The line 200*x** *+ 100*y* = 4000 meets the coordinate axes at *A*_{1}(20, 0) and *B*_{1}(0,40) respectively. By joining these points we obtain the line

200*x** *+ 100*y* = 4000.Clearly (0,0) does not satisfies the inequation* *200*x** *+ 100*y* ≥ 4000. So,the region in *xy* plane which does not contain the origin represents the solution set of the inequation* *200*x** *+ 100*y* ≥ 4000.

Region represented by *x** *+2*y* ≥ 50:

The line *x** *+2*y* = 50 meets the coordinate axes at *C*_{1}(50, 0) and *D*_{1}(0, 25) respectively. By joining these points we obtain the line*x** *+2*y* = 50.Clearly (0,0) does not satisfies the *x** *+2*y* ≥ 50. So,the region which does not contains the origin represents the solution set of the inequation *x** *+2*y* ≥ 50.

Region represented by 40*x** *+ 40*y* ≥ 1400:

The line 40*x** *+ 40*y* = 1400 meets the coordinate axes at *E*_{1}(35, 0) and *F*_{1}(0, 35) respectively. By joining these points we obtain the line

40*x** *+ 40*y** *= 1400.Clearly (0,0) does not satisfies the inequation 40*x** *+ 40*y* ≥ 1400. So,the region which does not contains the origin represents the solution set of the inequation 40*x** *+ 40*y* ≥ 1400.

Region represented by *x* ≥ 0 and* y* ≥ 0:

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations *x* ≥ 0, and *y *≥ 0.

The feasible region determined by the system of constraints 200*x** *+ 100*y* ≥ 4000,*x** *+2*y* ≥ 50, 40*x** *+ 40*y* ≥ 1400,* **x* ≥ 0, and *y* ≥ 0 are as follows.

The corner points of the feasible region are *B*_{1}(0, 40), *G*_{1}(5, 30), *H*_{1}(20, 15) and *C*_{1}(50, 0)

The value of the objective function at these points are given by the following table

Points | Value of Z |

B_{1} |
4(0)+3(40) = 120 |

G_{1} |
4(5)+3(30) = 110 |

H_{1} |
4(20) + 3(15) = 125 |

C_{1} |
4(50)+3(0) = 200 |

The minimum cost is Rs 110 which is at *G*_{1}(5, 30).

Hence, the required combination of food is 5 units of food *A* and 30 units of food *B*.