Question

consider the following linear programming problem: maximize $z = 2100x + 1300y$ (profit in $) given the following constraints: $3x + 2yleq18$, $x + 2yleq10$, $yleq4$, $x,ygeq0$. a. select the correct feasible region.

Explanation:

Step1: Analyze the first constraint

For $3x + 2y\leq18$, when $x = 0$, $y=9$ and when $y = 0$, $x = 6$. The line $3x + 2y=18$ has a negative - slope and the region satisfying $3x + 2y\leq18$ is below or on this line.

Step2: Analyze the second constraint

For $x + 2y\leq10$, when $x = 0$, $y = 5$ and when $y=0$, $x = 10$. The line $x + 2y=10$ has a negative - slope and the region satisfying $x + 2y\leq10$ is below or on this line.

Step3: Analyze the third constraint

The constraint $y\leq4$ represents the region below or on the horizontal line $y = 4$.

Step4: Consider non - negativity constraints

The constraints $x\geq0$ and $y\geq0$ mean the feasible region lies in the first quadrant.

Step5: Determine the feasible region

The intersection of all the regions defined by the constraints $3x + 2y\leq18$, $x + 2y\leq10$, $y\leq4$, $x\geq0$ and $y\geq0$ gives the feasible region.

Answer:

The feasible region is the polygon in the first - quadrant that is below or on the lines $3x + 2y=18$, $x + 2y=10$ and $y = 4$. (Based on the given graph, the shaded region in the first - quadrant bounded by the intersection of the lines corresponding to the constraints is the feasible region).