Question

The maximum value of \( z = 5x + 2y \) subject to the constraints: \[ x + y \leq 7, \quad x + 2y \leq 10, \quad x, y \geq 0 \]

• A. 10
• B. 26
• C. 35
• D. 70

Hint:

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

Answer 1

The Correct Option is C

Step 1: We are given the objective function \( z = 5x + 2y \) and the constraints: \[ x + y \leq 7, \quad x + 2y \leq 10, \quad x \geq 0, \quad y \geq 0. \] To find the maximum value of \( z \), we will first graph the constraints and identify the feasible region, and then evaluate the objective function at the corner points (vertices) of the feasible region. 
Step 2: Rewrite the constraints as equations: - \( x + y = 7 \) (Line 1), - \( x + 2y = 10 \) (Line 2). The feasible region is bounded by these lines and the axes. 
Step 3: Find the intersection points of these lines: 
- Intersection of \( x + y = 7 \) and \( x + 2y = 10 \): Solve the system of equations: \[ x + y = 7 \quad {(Equation 1)}, \] \[ x + 2y = 10 \quad {(Equation 2)}. \] From Equation 1, \( x = 7 - y \). 
Substitute into Equation 2: \[ (7 - y) + 2y = 10, \] \[ 7 + y = 10, \] \[ y = 3. \] Substitute \( y = 3 \) into \( x + y = 7 \): \[ x + 3 = 7 \quad \Rightarrow \quad x = 4. \] 
Thus, the intersection point is \( (4, 3) \). 
- Intersection of \( x + y = 7 \) and the x-axis (where \( y = 0 \)): \[ x + 0 = 7 \quad \Rightarrow \quad x = 7. \] 
Thus, the point is \( (7, 0) \). 
- Intersection of \( x + 2y = 10 \) and the y-axis (where \( x = 0 \)): \[ 0 + 2y = 10 \quad \Rightarrow \quad y = 5. \] 
Thus, the point is \( (0, 5) \). 
Step 4: Now, evaluate \( z = 5x + 2y \) at each corner point: - At \( (7, 0) \), \( z = 5(7) + 2(0) = 35 \), - At \( (4, 3) \), \( z = 5(4) + 2(3) = 20 + 6 = 26 \), - At \( (0, 5) \), \( z = 5(0) + 2(5) = 10 \). 
The maximum value of \( z \) is \( 35 \), which occurs at \( (7, 0) \).