Sum

Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs 5 per kg and rice costs Rs 4 per kg

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#### Solution

Let the cereal contain *x *kg of bran and *y* kg of rice.

Therefore,

\[x, y \geq 0\]

The given information can be tabulated as follows

Bran and rice contains at least 88 grams of protein and at least 36 milligrams of iron.

Thus,the constraints are

Protein(grams) | Iron(milligrams) | |

Bran | 80 | 40 |

Rice | 100 | 30 |

minimum requirement | 88 | 36 |

Bran and rice contains at least 88 grams of protein and at least 36 milligrams of iron.

Thus,the constraints are

\[80x + 100y \geq 88\]

\[40x + 30y \geq 36\]

\[40x + 30y \geq 36\]

It is given that bran costs Rs 5 per kg and rice costs Rs 4 per kg. Therefore, cost of

Hence, total profit is Rs (5

Let Z denote the total cost.

*x*kg of bran and*y*kg of rice is Rs 5*x**and Rs 4**y**respectively.*Hence, total profit is Rs (5

*x*+ 4*y*)Let Z denote the total cost.

∴ z = \[5x + 4y\]

Thus, the mathematical formulation of the given problem is Minimize \[Z = 5x + 4y\]

subject to

Thus, the mathematical formulation of the given problem is Minimize \[Z = 5x + 4y\]

subject to

\[80x + 100y \geq 88\]

\[40x + 30y \geq 36\]

First we will convert inequations into equations as follows :

80

Region represented by 80

The line 80

80

Region represented by 40

The line 40

Clearly (0,0) does not satisfies the inequation 40

Region represented by

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

The feasible region determined by the system of constraints 80

The feasible region determined by the system of constraints is

The corner points are

\[40x + 30y \geq 36\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :

80

*x*+ 100*y*= 60, 40*x*+ 30*y*= 36,*x*= 0 and*y*= 0Region represented by 80

*x*+ 100*y**≥ 88:*The line 80

*x*+ 100*y*= 60 meets the coordinate axes at*A*(1.1, 0) and*B*(0, 0.88)respectively. By joining these points we obtain the line 80*x*+ 100*y*= 60. Clearly (0,0) does not satisfies the*80**x*+ 100*y**≥ 88. So,the region which does not contains the origin represents the solution set of the inequation*80

*x*+ 100*y**≥ 88.*Region represented by 40

*x*+ 30*y*≥ 36:The line 40

*x*+ 30*y*= 36 meets the coordinate axes at*C*(0.9, 0) and*D*(0, 1.2) respectively. By joining these points we obtain the line 40*x*+ 30*y*= 36.Clearly (0,0) does not satisfies the inequation 40

*x*+ 30*y*≥ 36. So,the region which does not contains the origin represents the solution set of the inequation 40*x*+ 30*y*≥ 36.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 80

*x*+ 100*y**≥ 88, 40**x*+ 30*y*≥ 36,*x*≥ 0, and*y*≥ 0 are as follows.The feasible region determined by the system of constraints is

The corner points are

*D*(0, 1.2)*, E*(0.6, 0.4) and*A*(1.1, 0).The values of Z at these corner points are as follows:

Corner point | Z = 5x + 4y |

D(0, 1.2) |
4.8 |

E(0.6, 0.4) |
4.6 |

A(1.1, 0) |
5.5 |

The minimum value of Z is 4.6 which is attained at *E*(0.6, 0.4).

Hence, the minimum cost is Rs 4.6.

Concept: Graphical Method of Solving Linear Programming Problems

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