Which one of the options given represents the feasible region of the linear programming model:

                                        Maximize 45X₁, + 60X₂

                                                           X₁ ≤ 45

                                                          X₂ ≤ 50

                                                          10X₁ + 10X₂ ≥ 600

                                                          25X₁, + 5X₂, ≤  750 

The correct option is Region Q.

To determine which region represents the feasible region of the given linear programming problem, we analyze the constraints step-by-step:

Constraint 1:
$$X_1\le 45$$
This boundary is a vertical line at $X_1=45$. The inequality $X_1\le 45$ restricts the feasible region to the left side of this vertical line.

Constraint 2:
$$X_2\le 50$$
This boundary is a horizontal line at $X_2=50$. The inequality $X_2\le 50$ restricts the feasible region to be below this horizontal line.

Constraint 3:
$$10X_1+10X_2\ge 600$$
Dividing both sides by 10 gives:
$$X_1+X_2\ge 60$$
The boundary line is $X_1+X_2=60$ with intercepts at $(60,0)$ and $(0,60)$. Testing the origin $(0,0)$: $0+0\ge 60$ is False, which means the feasible region lies on the side opposite to the origin (above and to the right of the line).

Constraint 4:
$$25X_1+5X_2\le 750$$
Dividing both sides by 5 simplifies this to:
$$5X_1+X_2\le 150$$
The boundary line is $5X_1+X_2=150$ with intercepts at $(30,0)$ and $(0,150)$. Testing the origin $(0,0)$: $0\le 150$ is True, meaning the feasible region lies on the origin side of this boundary line (below and to the left).

Identifying the Feasible Region:
By combining these constraints, the feasible region is bounded by:
1. Below the horizontal line $X_2=50$
2. Above/right of the line $X_1+X_2=60$
3. Below/left of the line $5X_1+X_2=150$

Let's find the intersection points forming this bounded region:
• Intersection of $X_2=50$ and $X_1+X_2=60$:
$$X_1+50=60⟹X_1=10$$
This gives the vertex $(10,50)$.

• Intersection of $X_2=50$ and $5X_1+X_2=150$:
$$5X_1+50=150⟹5X_1=100⟹X_1=20$$
This gives the vertex $(20,50)$.

• Intersection of $X_1+X_2=60$ and $5X_1+X_2=150$:
Subtracting the first equation from the second:
$$4X_1=90⟹X_1=22.5$$
Then:
$$X_2=60-22.5=37.5$$
This gives the vertex $(22.5,37.5)$.

This forms a triangular region enclosed by the vertices $(10,50)$, $(20,50)$, and $(22.5,37.5)$. Comparing this to the provided graph in the image, this exact region is labeled as Region Q.