Question

A diet of two foods

F_{1}

and

F_{2}

contains nutrients thiamine, phosphorous and iron. The amount of each nutrient in each of the food (in milligrams per

25 gms

) is given in the following table:

Nutrients/Food	$F_{1}$	$F_{2}$
Thiamine	$0.25$	$0.10$
Phosphorous	$0.75$	$1.50$
Iron	$1.60$	$0.80$

The minimum requirement of the nutrients in the diet are

1.00 mg

of thiamine,

7.50 mg

of phosphorous and

10.00 mg

of iron. The cost of

F_{1}

is

20

paise per

25 gms

while the cost of

F_{2}

is

15

paise per

25 gms

. Find the minimum cost of diet.

Accepted Answer

Let $25 x$ grams of food $F_{1}$ and $25 y$ grams of food $F_{2}$ be used to fulfil the minimum requirement of thiamine, phosphorus and iron.

As, we are given

Nutrients/Food $F_{1}$ $F_{2}$

Thiamine $0.25$ $0.10$

Phosphorous $0.75$ $1.50$

Iron $1.60$ $0.80$

And the minimum requirement of the nutrients in the diet are $1.00 mg$ of thiamine, $7.50 mg$ of phosphorous and $10.00 mg$ of iron.

Therefore,

$0.25 x + 0.10 y \geq 1$

$0.75 x + 1.50 y \geq 7.5$

$1.6 x + 0.8 y \geq 10$

Since, the quantity cannot be negative

$∴ x, y \geq 0$

The cost of $F_{1}$ is $20$ paise per $25 gms$ while the cost of $F_{2}$ is $15$ paise per $25 gms$ . Therefore, the cost of $25 x$ grams of food $F_{1}$ and $25 y$ grams of food $F_{2}$ is

$Rs (0.20 x + 0.15 y) .$

Hence,

Minimize

$Z = 0.20 x + 0.15 y$

subject to

$0.25 x + 0.10 y \geq 1$

$0.75 x + 1.50 y \geq 7.5$

$1.6 x + 0.8 y \geq 10$

$x, y \geq 0$

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

$0.25 x + 0.10 y = 1, 0.75 x + 1.50 y = 7.5, 1.6 x + 0.8 y = 10, x = 0$ and $y = 0 .$

The line $0.25 x + 0.10 y = 1$ meets the coordinate axis at

$A (4, 0)$ and $B (0, 10)$ . Join these points to obtain the line $0.25 x + 0.10 y = 1$ .

Clearly, $(0, 0)$ does not satisfies the inequation $0.25 x + 0.10 y \geq 1$ . So, the region in $x y$ -plane that does not contains the origin represents the solution set of the given equation.

The line $0.75 x + 1.50 y = 7.5$ . meets the coordinate axis at $C (10, 0)$ and $D (0, 5)$ . Join these points to obtain the line $0.75 x + 1.50 y = 7.5$ .

Clearly, $(0, 0)$ does not satisfies the inequation $0.75 x + 1.50 y \geq 0.75$ . So, the region in $x y$ -plane that does not contains the origin represents the solution set of the given equation.

The line $1.6 x + 0.8 y = 10$ meets the coordinate axis at $E (\frac{25}{4}, 0)$ and $F (0, \frac{25}{2})$ Join these points to obtain the line $1.6 x + 0.8 y = 10$ .

Clearly, $(0, 0)$ does not satisfies the inequation $1.6 x + 0.8 y \geq 10$ . So, the region in $x y$ -plane that does not contains the origin represents the solution set of the given equation.

Region represented by $x \geq 0$ and $y \geq 0$ :

Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations.

These lines are drawn using a suitable scale.

The corner points of the feasible region are $F (0, 12.5), G (5, 2.5), C (10, 0)$

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

Points Value of $Z$

$F$ $0.20 (0) + 0.15 (12.5) = 1.875$

$G$ $0.20 (5) + 0.15 (2.5) = 1.375$

$C$ $0.20 (10) + 0.15 (0) = 2$

Thus, the minimum cost is at $G$ which is $Rs 1.375$ .

R D Sharma Solutions for Chapter: Linear Programming, Exercise 3: EXERCISE

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Linear Programming, Exercise 3: EXERCISE

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