Question

A hospital dietician wishes to find the cheapest combination of two foods, A and B, that contains at least 0.5 milligram of thiamin and at least 600 calories. Each unit of A contains 0.12 milligram of thiamin and 100 calories, while each unit of B contains 0.10 milligram of thiamin and 150 calories. If each food costs 10 paise per unit, how many units of each should be combined at a minimum cost?

Answer 1

Let the dietician wishes to mix x units of food A and y units of food B.
Therefore, $x,y\ge 0$

The given information can be tabulated as follows

	Thiamine(mg)	Calories
Food A	0.12	100
Food B	0.1	150
Minimum requirement	0.5	600

According to the question,

The constraints are
$$\begin{gathered} 0.12x+0.1y\ge 0.5 \\ 100x+150y\ge 600 \end{gathered}$$

It is given that each food costs 10 paise per units

Therefore,
Total cost, Z = $10x+10y$

Thus, the mathematical formulat​ion of the given linear programmimg problem is

$$\begin{gathered} 0.12x+0.1y\ge 0.5 \\ 100x+150y\ge 600 \end{gathered}$$

Region represented by 0.12x +0.1y ≥ 0.5:
The line 0.12x + 0.6y = 20 meets the coordinate axes at $A_1\left(\frac{25}{6},0\right)$ and $B_1\left(0,5\right)$ respectively. By joining these points we obtain the line
0.12x + 0.6y = 20.Clearly (0,0) does not satisfies the 0.12x + 0.6y = 20. So,the region which does not contains the origin represents the solution set of the inequation 0.12x +0.1y ≥ 0.5.

Region represented by 100x + 150y ≥ 600:
The line 100x + 150y = 600 meets the coordinate axes at $C_1\left(6,0\right)$ and $D_1\left(0,4\right)$ respectively. By joining these points we obtain the line 100x + 150y = 600. Clearly (0,0) does not satisfies the inequation 100x + 150y ≥ 600. So,the region which does not contains the origin represents the solution set of the inequation 100x + 150y ≥ 600.

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 0.12x +0.1y ≥ 0.5, 100x + 150y ≥ 600, x ≥ 0, and y ≥ 0 are as follows.



The corner points are B₁(0, 5), $E_1\left(\frac{15}{8},\frac{11}{4}\right)$, $C_1\left(6,0\right)$

The values of Z at these corner points are as follows

Corner point	Z= 10x +10y
B1	50
E1	46.2
C1	60

The minimum value of Z is at $E_1\left(\frac{15}{8},\frac{11}{4}\right)$.

Hence, cheapest combination of foods will be 1.875 units of food A and 2.75 units of food B.