# 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 - Mathematics

Sum

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 Acontains 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?

#### Solution

Let the dietician wishes to mix units of food A and y units of food B.
Therefore,

$x, y \geq 0$
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

$0 . 12x + 0 . 1y \geq 0 . 5$
$100x + 150y \geq 600$

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

$0 . 12x + 0 . 1y \geq 0 . 5$
$100x + 150y \geq 600$

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 ≥ 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.

Concept: Graphical Method of Solving Linear Programming Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.3 | Q 4 | Page 39