Question

Max of \( Z = 7x + 10y \) subject to \( x + y \geq 3 \), \( x + 2y \geq 4 \), \( x, y \geq 0 \).

Hint:

In linear programming, the maximum or minimum of a linear function occurs at one of the vertices of the feasible region.

Answer 1

Step 1: Plot the constraints.
We are given the following constraints: 1. \( x + y \geq 3 \) 2. \( x + 2y \geq 4 \) 3. \( x \geq 0, \, y \geq 0 \) These represent linear inequalities. We first solve for the boundaries of the inequalities.
Step 2: Find the points of intersection.
From the first constraint \( x + y = 3 \), we get the line \( x = 3 - y \). From the second constraint \( x + 2y = 4 \), we get the line \( x = 4 - 2y \). To find the points of intersection, solve the system of equations: \[ x + y = 3 \] \[ x + 2y = 4 \] Subtract the first equation from the second: \[ x + 2y - (x + y) = 4 - 3 \] \[ y = 1 \] Substitute \( y = 1 \) into \( x + y = 3 \): \[ x + 1 = 3 \quad \Rightarrow \quad x = 2 \] Thus, the intersection point is \( (2, 1) \).
Step 3: Evaluate \( Z \) at the vertices.
The feasible region is formed by the constraints. The vertices are the points of intersection of the lines and the axes. We now evaluate \( Z = 7x + 10y \) at the points \( (0, 3) \), \( (2, 1) \), and \( (0, 4) \). - At \( (0, 3) \): \[ Z = 7(0) + 10(3) = 30 \] - At \( (2, 1) \): \[ Z = 7(2) + 10(1) = 14 + 10 = 24 \] - At \( (0, 4) \): \[ Z = 7(0) + 10(4) = 40 \]
Step 4: Conclusion.
The maximum value of \( Z \) is 40 at \( (0, 4) \).