Question

A manufacture is to market a new fertilizer which is to be a mixture of ingredients A and B. The properties of the two ingredients are as give on the table below
Properties of the ingredients Cost per kg
Bone meal Nitrogen Lime Phosphate
Ingredients A 20% 30% 40% 10% 12
B 40% 10% 50% 50% 8

It has been decided that the fertilizer will:
i. Be sold in bags containing 100 kilograms
ii. Be containing at least 15 kilograms of nitrogen
iii. Must contain at least 8 kilograms of phosphate
iv. Must contain at least 25 kilograms of bone meal
The manufacturer wants to meet the above requirements at minimum cost possible.
Required
Develop a linear program and hence find the optimal solution for the problem.

Answer 1

Answer on Question #51725 - Programming | Other

Manufacture is to market a new fertilizer which is to be a mixture of ingredients A and B. The properties of the two ingredients are as given on the table below

It has been decided that the fertilizer will:

i. Be sold in bags containing 100 kilograms

ii. Be containing at least 15 kilograms of nitrogen

iii. Must contain at least 8 kilograms of phosphate

iv. Must contain at least 25 kilograms of bone meal

The manufacturer wants to meet the above requirements at minimum cost possible.

Required Develop a linear program and hence find the optimal solution for the problem.

Solution:

Since the exercise is asking for the number of kilograms of each ingredient required for the mixture, our variables will stand for the number of kilograms of each:

$x$ - kgs of ingredient A, and $y$ - kgs of ingredient B.

The optimization equation will be the cost relation `C = 12 * x + 8 * y`.

Subject to:

i. `x + y = 100` (Weight Constraint)

ii. `0.3x + 0.1y >= 15` (Nitrogen Constraint)

iii. `0.1x + 0.05y >= 8` (Phosphate Constraint)

iv. `0.2x + 0.4y >= 25` (Bone Meal Constraint)

Since we can't use negative amounts of either fertilizer: `x >= 0`, `y >= 0`.

Remove the inequalities and solve thus:

i. `x + y = 100` If `x = 0` then `y = 100` and If `y = 0` then `x = 100`.

ii. `0.3x + 0.1y = 15` If `x = 0` then `y = 15 / 0.1 = 150` and If `y = 0` then `x = 15 / 0.3 = 50`

iii. `0.1x + 0.05y = 8` If `x = 0` then `y = 8 / 0.05 = 160` and If `y = 0` then `x = 8 / 0.1 = 80`

iv. `0.2x + 0.4y = 25` If `x = 0` then `y = 25 / 0.4 = 62.5` and If `y = 0` then `x = 25 / 0.2 = 125`

The above computations are then graphed as follows:

In this graph we could take points from feasible region that lie on the dotted dark blue line.

Reading from the graph:

The above implies that the best combination of ingredients that would minimize total cost is 60 kg of A and 40 kg of B.

Answer: 60 kg of A and 40 kg of B.

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