Solution

The correct answer is Rs. 133.3

Explanation:

To find the maximum total income, we need to solve the linear programming problem formulated based on the given constraints and objective function.

Objective function: \(\text{Maximize } Z = 30x + 50y\)

Constraints:

• \(2x + 3y \leq 8\) Red wool constraint
• \(3x + 2y \leq 10\) Green wool constraint
• \(2y \leq 15\) Blue wool constraint, which simplifies to \(y \leq 7.5\)
• \(x \geq 0\)
• \(y \geq 0\)

First, let's graph these constraints and identify the feasible region.

For \(2x + 3y \leq 8\) :

When x = 0 : 3y = 8 so \(y = \frac{8}{3}\)
When y = 0 : 2x = 8 so x = 4
For \(3x + 2y \leq 10\) :

When x = 0 : 2y = 10 so y = 5
When y = 0 : 3x = 10 so \(x = \frac{10}{3}\)
For \(y \leq 7.5\) :

This line is a horizontal line at y = 7.5
Now we need to find the intersection points of these lines to determine feasible points.

Intersection of 2x + 3y = 8 and 3x + 2y = 10 :

Solve the system: 2x + 3y = 8 3x + 2y = 10

Multiply the first equation by 3: 6x + 9y = 24

Multiply the second equation by 2: 6x + 4y = 20

Subtract the second equation from the first: \(5y = 4 \rightarrow y = \frac{4}{5}\)

Substitute \(y = \frac{4}{5}\) into 2x + 3y = 8:

• \(2x + 3 \left(\frac{4}{5}\right) = 8\)
• \(2x + \frac{12}{5} = 8 \)
• \(2x = 8 - \frac{12}{5}\)
• \( 2x = \frac{40}{5} - \frac{12}{5}\)
• \(2x = \frac{28}{5}\)
•   \(x = \frac{14}{5}\)

Therefore, the intersection point is \left \frac{14}{5}, \frac{4}{5} \right .

Other boundary points to consider:

• (0, 0)
• (\(0, \frac{8}{3}\))
• (0, 5)
• (4, 0)
• \(\left( \frac{10}{3}, 0 \right)\)

Finally, evaluate the objective function Z = 30x + 50y at these points to find the maximum income.

At (0, 0) : \(Z = 30(0) + 50(0) = 0\)

At \((0, \frac{8}{3})\) : \(Z = 30(0) + 50\left(\frac{8}{3}\right) = 50 \times \frac{8}{3} = \frac{400}{3} \approx 133.33\)

At (0, 5) : \(Z = 30(0) + 50(5) = 250\) The point (0, 5)  is not a feasible point, it lies outside considering 3x + 2y = 10

At (4, 0) : \(Z = 30(4) + 50(0) = 120\)

At \( \left( \frac{10}{3}, 0 \right)\): \( Z = 30 \left( \frac{10}{3} \right) + 50(0) = 100\)

At \(\left( \frac{14}{5}, \frac{4}{5} \right)\) : Z = \(Z = 30 \left( \frac{14}{5} \right) + 50 \left( \frac{4}{5} \right) = 84 + 40 = 124\)

Therefore, the maximum total income is at \((0, \frac{8}{3})\) and is equal to \(\approx 133.33\) .

So the correct answer is: Rs. 133.3