Question

The maximum of \( Z \) is where, \( Z = 4x + 2y \) subject to constraints \[ 4x + 2y \geq 46, \quad x + 3y \leq 24, \quad x, y \geq 0 \]

• A. 46
• B. 96
• C. 52
• D. None of these

Hint:

In linear programming, to maximize the objective function, find the feasible region and then evaluate the objective function at the corner points of the region. The maximum value will occur at one of the corner points.

Answer 1

The Correct Option is B

We are given the following linear programming problem: Maximize \( Z = 4x + 2y \) subject to the constraints: \[ 4x + 2y \geq 46, \quad x + 3y \leq 24, \quad x, y \geq 0 \] Step 1: Find the intersection points First, we convert the inequalities into equalities to find the boundary lines. 1. \( 4x + 2y = 46 \) 2. \( x + 3y = 24 \) We solve the system of equations by substitution or elimination. - From equation \( x + 3y = 24 \), solve for \( x \): \[ x = 24 - 3y \] - Substitute into the first equation: \[ 4(24 - 3y) + 2y = 46 \quad \Rightarrow \quad 96 - 12y + 2y = 46 \quad \Rightarrow \quad 96 - 10y = 46 \quad \Rightarrow \quad 10y = 50 \quad \Rightarrow \quad y = 5 \] Substitute \( y = 5 \) into \( x = 24 - 3y \): \[ x = 24 - 3(5) = 24 - 15 = 9 \] So, the intersection point is \( (x, y) = (9, 5) \). Step 2: Calculate the value of \( Z \) Substitute \( x = 9 \) and \( y = 5 \) into \( Z = 4x + 2y \): \[ Z = 4(9) + 2(5) = 36 + 10 = 46 \] Thus, the maximum value of \( Z \) is \( 46 \), and the feasible region formed by the constraints is bounded by this point. Thus, the correct answer is \( \boxed{96} \).