Question

Maximum value of \( z = 12x + 3y \), subject to constraints \( x \geq 0, y \geq 0, x + y \leq 5 \) and \( 3x + y \leq 9 \) is

• A. 15
• B. 36
• C. 60
• D. 40

Hint:

In linear programming problems, always plot the constraints to identify the feasible region and check the values at the vertices of the region.

Answer 1

The Correct Option is B

The given constraints are: 1. \( x \geq 0 \) 2. \( y \geq 0 \) 3. \( x + y \leq 5 \) 4. \( 3x + y \leq 9 \) We need to find the maximum value of \( z = 12x + 3y \). First, plot the constraints and identify the feasible region. The lines corresponding to the inequalities are: \[ x + y = 5 \quad \text{and} \quad 3x + y = 9 \] Now, find the intersection points of these lines: 1. Intersection of \( x + y = 5 \) and \( 3x + y = 9 \): \[ x + y = 5 \quad \text{and} \quad 3x + y = 9 \] Subtract the first equation from the second: \[ (3x + y) - (x + y) = 9 - 5 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = 2 \] Substitute \( x = 2 \) in \( x + y = 5 \): \[ 2 + y = 5 \quad \Rightarrow \quad y = 3 \] So the intersection point is \( (2, 3) \). Now calculate \( z \) at this point: \[ z = 12x + 3y = 12(2) + 3(3) = 24 + 9 = 36 \] Thus, the maximum value of \( z \) is 36.