Question

The maximum value of $ Z = 12x + 13y $, subject to constraints $ x \geq 0, \quad y \geq 0, \quad x + y \leq 5, \quad 3x + y \leq 9 $ is

• A. 63
• B. 65
• C. 60
• D. 117

Hint:

When solving linear programming problems, always evaluate the objective function at the vertices of the feasible region to find the optimal solution.

Answer 1

The Correct Option is B

We are given the objective function \( Z = 12x + 13y \), and we need to maximize it subject to the following constraints: 
1. \( x \geq 0 \) 
2. \( y \geq 0 \) 
3. \( x + y \leq 5 \) 
4. \( 3x + y \leq 9 \) 
To solve this, we graph the constraints and find the feasible region formed by these inequalities. The vertices of the feasible region are where the lines intersect, as these are the points that will give the maximum or minimum value of \( Z \). 
Step 1: Graph the constraints
The first constraint is \( x \geq 0 \), which represents all points to the right of the \( y \)-axis. The second constraint is \( y \geq 0 \), which represents all points above the \( x \)-axis. The third constraint is \( x + y \leq 5 \), which is the region below the line \( x + y = 5 \). The fourth constraint is \( 3x + y \leq 9 \), which is the region below the line \( 3x + y = 9 \). 
Step 2: Find the intersection points
We need to find the points where the constraint lines intersect, as these will be the vertices of the feasible region. 1. Intersection of \( x + y = 5 \) and \( 3x + y = 9 \): Solve the system of equations: \[ x + y = 5 \quad \text{(1)} \] \[ 3x + y = 9 \quad \text{(2)} \] Subtract equation (1) from equation (2): \[ (3x + y) - (x + y) = 9 - 5 \] \[ 2x = 4 \quad \Rightarrow \quad x = 2 \] Substitute \( x = 2 \) into equation (1): \[ 2 + y = 5 \quad \Rightarrow \quad y = 3 \] So, the intersection point is \( (2, 3) \). 2. Intersection of \( x + y = 5 \) with \( x = 0 \) (y-axis): Substitute \( x = 0 \) into the equation \( x + y = 5 \): \[ 0 + y = 5 \quad \Rightarrow \quad y = 5 \] So, the point is \( (0, 5) \). 3. Intersection of \( 3x + y = 9 \) with \( y = 0 \) (x-axis): Substitute \( y = 0 \) into \( 3x + y = 9 \): \[ 3x = 9 \quad \Rightarrow \quad x = 3 \] So, the point is \( (3, 0) \). 
Step 3: Calculate \( Z \) at the vertices
We now calculate the value of \( Z = 12x + 13y \) at each vertex: - At \( (2, 3) \): \[ Z = 12(2) + 13(3) = 24 + 39 = 63 \] - At \( (0, 5) \): \[ Z = 12(0) + 13(5) = 0 + 65 = 65 \] - At \( (3, 0) \): \[ Z = 12(3) + 13(0) = 36 + 0 = 36 \] Thus, the maximum value of \( Z \) is 65 at the point \( (0, 5) \).