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A Diet is to Contain at Least 80 Units of Vitamin a and 100 Units of Minerals. Two Foods F1 and F2 Are Available - Mathematics

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Sum

A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods F1and F2 are available. Food F1 costs Rs 4 per unit and F2 costs Rs 6 per unit one unit of food F1 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem and find graphically the minimum cost for diet that consists of mixture of these foods and also meets the mineral nutritional requirements

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Solution

Let the dietician wishes to mix units of food F1 and y units of food F2.
Clearly, \[x, y \geq 0\]

The given information can be tabulated as follows:

  Vitamin A Minerals
Food F1 3 4
 Food F2 6 3
Minimum requirement 80 100

The constraints are

\[3x + 6y \geq 80\] 
\[4x + 3y \geq 100\]

It is given that cost of food F1 and F2 is Rs 4 and Rs 6 per unit respectively. Therefore, cost of units of food F1 and y units of food F2 is Rs 4x  and Rs 6y respectively.
Let Z denote the total cost
∴ Z = 4x + 6y 
Thus, the mathematical formulat‚Äčion of the given linear programmimg problem is 
Minimize

\[Z = 4x + 6y\] subject to 

\[3x + 6y \geq 80\]
\[4x + 3y \geq 100\]

\[x, y \geq 0\]

First, we will convert the given inequations into equations, we obtain the following equations:
3x + 6y = 80, 4x + 3y = 100= 0 and y = 0

The line 3x + 6y = 80 meets the coordinate axis at

\[A\left( \frac{80}{3}, 0 \right)\] and\[B\left( 0, \frac{40}{3} \right)\] Join these points to obtain the line 3x + 6y = 80. Clearly, (0, 0) does not satisfies the inequation 3x + 6y ≥ 80. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.
The line 4x + 3y = 100 meets the coordinate axis at C(25, 0) and \[D\left( 0, \frac{100}{3} \right)\]  Join these points to obtain the line 4x + 3y = 100. Clearly, (0, 0) does not satisfies the inequation 4x + 3y ≥ 100. So, the region in xy-plane that does not contains the origin represents the solution set of the given equation.

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.
These lines are drawn using a suitable scale.

The feasible region determined by the system of constraints is 

The corner points are D

\[\left( 0, \frac{100}{3} \right)\] , E \[\left( 24, \frac{4}{3} \right)\] and \[A\left( \frac{80}{3}, 0 \right)\] The values of Z at these corner points are as follows:
 

 

Corner point

Z= 4x + 6y
\[\left( 0, \frac{100}{3} \right)\] 200
\[\left( 24, \frac{4}{3} \right)\] 104
\[A\left( \frac{80}{3}, 0 \right)\]
\[\frac{320}{3}\]
 

The minimum value of Z is Rs 104 which is at \[\left( 24, \frac{4}{3} \right)\]

 
Concept: Graphical Method of Solving Linear Programming Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.3 | Q 6 | Page 39

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