Question

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

Explanation:

Step1: Rewrite inequalities as equations

Rewrite $3x + 2y=18$, $x + 2y=10$, $y = 4$, $x=0$, $y = 0$.

Step2: Find intersection points of boundary - lines

For $3x + 2y=18$ and $x=0$, substitute $x = 0$ into $3x + 2y=18$:
$2y=18$, so $y = 9$. Inter - section point is $(0,9)$.
For $3x + 2y=18$ and $y = 0$, substitute $y = 0$ into $3x + 2y=18$:
$3x=18$, so $x = 6$. Inter - section point is $(6,0)$.
For $x + 2y=10$ and $x=0$, substitute $x = 0$ into $x + 2y=10$:
$2y=10$, so $y = 5$. Inter - section point is $(0,5)$.
For $x + 2y=10$ and $y = 0$, substitute $y = 0$ into $x + 2y=10$:
$x=10$. Inter - section point is $(10,0)$.
Solve the system of equations

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

Subtract the second equation from the first: $(3x + 2y)-(x + 2y)=18 - 10$
$2x=8$, so $x = 4$. Substitute $x = 4$ into $x + 2y=10$, we get $4+2y=10$, $2y=6$, $y = 3$. Inter - section point is $(4,3)$.
Solve the system of equations

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

Substitute $y = 4$ into $x + 2y=10$, we get $x+2\times4=10$, $x = 2$. Inter - section point is $(2,4)$.
Solve the system of equations

$$\begin{cases}3x + 2y=18\\y = 4\end{cases}$$

Substitute $y = 4$ into $3x + 2y=18$, we get $3x+2\times4=18$, $3x=10$, $x=\frac{10}{3}$. Inter - section point is $(\frac{10}{3},4)$.

Step3: Determine the feasible region

The inequalities $x\geq0$, $y\geq0$ mean the region is in the first - quadrant.
The inequality $3x + 2y\leq18$ represents the region below the line $3x + 2y=18$.
The inequality $x + 2y\leq10$ represents the region below the line $x + 2y=10$.
The inequality $y\leq4$ represents the region below the line $y = 4$.
The feasible region is a polygon bounded by the intersection points of the boundary lines that satisfy all the inequalities.

Since no options for the feasible - region are given, we can describe it in terms of its vertices. The vertices of the feasible region are $(0,0)$, $(0,4)$, $(2,4)$, $(4,3)$, $(6,0)$.

Answer:

The feasible region is a polygon in the first - quadrant with vertices $(0,0)$, $(0,4)$, $(2,4)$, $(4,3)$, $(6,0)$ (assuming no options are provided for selection). If options were given, we would check which one has these vertices or is bounded by the correct lines.