Question

Maximize \( z = 5x + 2y \) subject to \( 2x + y \leq 8 \), \( x \geq 0 \), \( y \geq 0 \). What is the maximum value of \( z \)?

• A. \( 20 \)
• B. \( 30 \)
• C. \( 25 \)
• D. \( 40 \)

Hint:

For linear programming, maximum occurs at vertices of the feasible region.

Answer 1

The Correct Option is A

The problem involves maximizing the objective function \( z = 5x + 2y \) subject to the constraints \( 2x + y \leq 8 \), with \( x \geq 0 \) and \( y \geq 0 \).

Step 1: Identify the constraints and the feasible region.
We rewrite the constraint \( 2x + y \leq 8 \) as the line equation \( y = -2x + 8 \). The feasible region is bounded by this line, the x-axis, and the y-axis, forming a triangle with vertices at the origin (0,0), a point on the x-axis, and a point on the y-axis.

Step 2: Find the intersection points with the axes.
For the x-axis \( (y = 0) \), solve \( 2x = 8 \) to get \( x = 4 \). Thus, the point is (4, 0).
For the y-axis \( (x = 0) \), solve \( y = 8 \). Thus, the point is (0, 8).

Step 3: Evaluate the objective function at each vertex.
1. At (0, 0): \( z = 5(0) + 2(0) = 0 \).
2. At (4, 0): \( z = 5(4) + 2(0) = 20 \).
3. At (0, 8): \( z = 5(0) + 2(8) = 16 \).

The maximum value of \( z \) at these points is 20. 
Thus, the maximum value that \( z \) can reach in the feasible region given the constraints is 20.

Vertex	\((x, y)\)	Value of \(z\)
1	(0, 0)	0
2	(4, 0)	20
3	(0, 8)	16