Let the two variables $ x $ and $ y $ satisfy the following conditions:
\[
x + y \leq 50, \quad x + 2y \leq 80, \quad 2x + y \geq 20, \quad x, y \geq 0.
\]
Then the maximum value of $ Z = 4x + 3y $ is:
Show Hint
- 120
- 170
- 200
- 210
The Correct Option is C
Solution and Explanation
We are given a system of linear inequalities:
1. \( x + y \leq 50 \), 2. \( x + 2y \leq 80 \), 3. \( 2x + y \geq 20 \), 4. \( x, y \geq 0 \). We are asked to maximize \( Z = 4x + 3y \).
Step 1: Graph the constraints.
We graph the inequalities to find the feasible region.
Step 2: Identify the corner points.
By solving the system of equations, we find the corner points of the feasible region.
Step 3: Calculate \( Z \) at each corner point.
For each corner point, we calculate the value of \( Z = 4x + 3y \). The maximum value occurs at \( x = 40 \) and \( y = 10 \), giving: \[ Z = 4(40) + 3(10) = 160 + 40 = 200. \] Thus, the maximum value of \( Z \) is \( 200 \).
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