5.3 Q-7

Question Statement

Each unit of food X costs Rs. 25 and contains 2 units of protein and 4 units of iron. Each unit of food Y costs Rs. 30 and contains 3 units of protein and 2 units of iron. Each animal must receive at least 12 units of protein and 16 units of iron each day. The goal is to determine how many units of each food should be fed to each animal at the smallest possible cost.

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Background and Explanation

In this problem, we are dealing with a linear programming optimization problem. We need to find the least cost combination of food units X and Y that satisfy the nutritional requirements for protein and iron. The problem involves:

• A cost function for food items X and Y.
• Constraints on the required units of protein and iron.
• The goal is to minimize the cost while fulfilling these constraints.

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Solution

1. Define Variables and the Cost Function

Let:

• x = number of units of food X.
• y = number of units of food Y.

The cost function (to minimize) is:

$$f(x, y) = 25x + 30y$$

2. Set up the Constraints

The animal must receive:

• At least 12 units of protein, which gives the constraint: $3x + 3y \geq 12$
• At least 16 units of iron, which gives the constraint: $4x + 2y \geq 16$
• Non-negativity constraints: $x \geq 0, \quad y \geq 0$

3. Graph the Constraints

First constraint:

$3x + 3y = 12$

• When $x = 0$, $y = 4$. The point $(0, 4)$ lies on the line.
• When $y = 0$, $x = 4$. The point $(4, 0)$ lies on the line.

Second constraint:

$4x + 2y = 16$

• When $x = 0$, $y = 8$. The point $(0, 8)$ lies on the line.
• When $y = 0$, $x = 4$. The point $(4, 0)$ lies on the line.

4. Find the Intersection Points

The feasible region lies in the first quadrant and is formed by the intersection of the constraints. By solving the system of equations for the points of intersection:

• From $3x + 3y = 12$, we solve for $y$: $y = \frac{12 - 3x}{3}$

• Substituting this into $4x + 2y = 16$: $4x + 2\left(\frac{12 - 3x}{3}\right) = 16$ Simplifying: $4x + \frac{24 - 6x}{3} = 16$ $12x + 24 - 6x = 48$ $6x = 24$ $x = 4$

• Substituting $x = 4$ into $3x + 3y = 12$: $3(4) + 3y = 12 \Rightarrow 12 + 3y = 12 \Rightarrow y = 0$

Thus, the corner points are $(3, 2), (6, 0), (0, 8)$.

5. Evaluate the Cost at Each Corner Point

Now, evaluate the cost function at the corner points:

• At $(3, 2)$: $f(3, 2) = 25(3) + 30(2) = 75 + 60 = 135$

• At $(6, 0)$: $f(6, 0) = 25(6) + 30(0) = 150$

• At $(0, 8)$: $f(0, 8) = 25(0) + 30(8) = 240$

6. Conclusion

The minimum cost occurs at the corner point $(3, 2)$, where the cost function value is 135 Rs. Hence, the optimal solution is to feed 3 units of food X and 2 units of food Y to minimize the cost.

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Key Formulas or Methods Used

• Cost Function: $f(x, y) = 25x + 30y$

• Constraints:

  • $3x + 3y \geq 12$ (Protein constraint)
  • $4x + 2y \geq 16$ (Iron constraint)
  • $x \geq 0, y \geq 0$ (Non-negativity constraints)

• Linear Programming Method: Graphing and solving the system of inequalities to find feasible corner points, then evaluating the cost function at each corner point.

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Summary of Steps

1. Define the variables and the cost function.
2. Set up the constraints based on the problem description.
3. Graph the constraints to find the feasible region.
4. Solve for the intersection points of the constraints.
5. Evaluate the cost function at each corner point of the feasible region.
6. Determine the corner point that gives the minimum cost.