Question

Find the maximum value of Z = 8x + 5y under the constraints \( 5x + 3y \le 15 \), \( 2x + 5y \le 10 \) and \( x \ge 0, y \ge 0 \) by graphical method.

Hint:

When solving an LPP, carefully plot the lines and shade the feasible region. The corner points are the origin, the intercepts, and the intersections of the boundary lines that lie within the feasible region. Always check all corner points.

Answer 1

Step 1: Understanding the Concept:
This is a Linear Programming Problem (LPP). The solution involves finding the feasible region determined by the linear constraints and then evaluating the objective function Z at the vertices (corner points) of this region. The maximum value of Z will occur at one of these vertices.
Step 2: Key Formula or Approach:
1. Plot the boundary lines of the inequalities.
2. Identify the feasible region by finding the common area satisfying all constraints.
3. Determine the coordinates of the corner points of the feasible region.
4. Substitute the coordinates of each corner point into the objective function Z.
5. The largest value obtained is the maximum value of Z.
Step 3: Detailed Explanation or Calculation:
The constraints are: 1. \( 5x + 3y \le 15 \) 2. \( 2x + 5y \le 10 \) 3. \( x \ge 0, y \ge 0 \) The feasible region is in the first quadrant.
Corner Points:
- O(0, 0): The origin.
- A: The x-intercept of \( 5x + 3y = 15 \). Let \( y=0 \), then \( 5x=15 → x=3 \). Point A is (3, 0).
- C: The y-intercept of \( 2x + 5y = 10 \). Let \( x=0 \), then \( 5y=10 → y=2 \). Point C is (0, 2).
- B: The intersection of \( 5x + 3y = 15 \) and \( 2x + 5y = 10 \).
Multiply the first equation by 5 and the second by 3:
\( 25x + 15y = 75 \)
\( 6x + 15y = 30 \)
Subtract the second from the first: \( 19x = 45 → x = \frac{45}{19} \).
Substitute x back into \( 2x + 5y = 10 \):
\( 2(\frac{45}{19}) + 5y = 10 → \frac{90}{19} + 5y = 10 → 5y = 10 - \frac{90}{19} = \frac{190-90}{19} = \frac{100}{19} → y = \frac{20}{19} \).
Point B is \( (\frac{45}{19}, \frac{20}{19}) \).
Evaluate Z at corner points:
- At O(0, 0): \( Z = 8(0) + 5(0) = 0 \)
- At A(3, 0): \( Z = 8(3) + 5(0) = 24 \)
- At C(0, 2): \( Z = 8(0) + 5(2) = 10 \)
- At B(\( \frac{45}{19}, \frac{20}{19} \)): \( Z = 8(\frac{45}{19}) + 5(\frac{20}{19}) = \frac{360}{19} + \frac{100}{19} = \frac{460}{19} \approx 24.21 \)
Step 4: Final Answer:
Comparing the values of Z at the corner points (0, 24, 10, and 460/19), the maximum value is \( \frac{460}{19} \).