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: Rewrite inequalities as equations

Rewrite $3x + 2y=18$, $x + 2y=10$, $y = 4$, $x=0$, $y = 0$ to find boundary - lines.

Step2: Find x - intercepts

For $3x + 2y=18$, when $y = 0$, $x = 6$; for $x + 2y=10$, when $y = 0$, $x = 10$.

Step3: Find y - intercepts

For $3x + 2y=18$, when $x = 0$, $y = 9$; for $x + 2y=10$, when $x = 0$, $y = 5$.

Step4: Find intersection points

Solve the system of equations

$$\begin{cases}3x + 2y=18\\x + 2y=10\end{cases}$$

. Subtract the second equation from the first: $(3x + 2y)-(x + 2y)=18 - 10$, $2x=8$, $x = 4$. Substitute $x = 4$ into $x + 2y=10$ gives $4+2y=10$, $2y=6$, $y = 3$.
Solve the system of equations

$$\begin{cases}x + 2y=10\\y = 4\end{cases}$$

, substitute $y = 4$ into $x + 2y=10$ gives $x+2\times4=10$, $x = 2$.

Step5: Determine the feasible region

The inequalities $x\geq0$, $y\geq0$ mean the region is in the first - quadrant. The feasible region is bounded by the intersection of $3x + 2y\leq18$, $x + 2y\leq10$, $y\leq4$, $x\geq0$ and $y\geq0$. It is a polygon with vertices $(0,0)$, $(0,4)$, $(2,4)$, $(4,3)$ and $(6,0)$.

(Note: Since the graph in the question is incomplete, we can't directly select from it based on the information provided. But the above steps show how to find the feasible - region.)

Answer:

The feasible region is a polygon in the first - quadrant with vertices $(0,0)$, $(0,4)$, $(2,4)$, $(4,3)$ and $(6,0)$.