Question:medium

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

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

Updated On: Nov 26, 2025
  • \( 20 \)
  • \( 30 \)
  • \( 25 \)
  • \( 40 \)
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The Correct Option is A

Solution and Explanation

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

Step 1: Define constraints and identify the feasible region.
The constraint \( 2x + y \leq 8 \) is represented by the line \( y = -2x + 8 \). The feasible region is enclosed by this line, the x-axis, and the y-axis, forming a triangular area with vertices at the origin (0,0), an x-intercept, and a y-intercept.

Step 2: Determine the intersection points with the coordinate axes.
For the x-axis (\(y = 0\)), solving \( 2x = 8 \) yields \( x = 4 \). The point is (4, 0).
For the y-axis (\(x = 0\)), \( y = 8 \). The point is (0, 8).

Step 3: Calculate the objective function's value 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 highest value of \( z \) among these vertices is 20.
Therefore, the maximum value \( z \) can attain within the defined feasible region is 20.

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