Question

The solution set of the inequalities $4x+3y\le60$, $y\ge2x$, $x\ge3$, $x, y\ge0$ is represented by region

• A. $S_{2}$ region
• B. $S_{1}$ region
• C. $S_{3}$ region
• D. $S_{4}$ region

Hint:

Graphing Tip:When dealing with multiple inequalities, find a test coordinate $(x,y)$ that visibly sits in the bounded central area to quickly verify the correct overlapping region.The intersection of $x=3$ and $y=2x$ provides a sharp corner vertex at $(3,6)$ which can help orient the graph.

Answer 1

The Correct Option is A

Concept:
Linear Programming - Graphical Solution of Linear Inequalities.

Step 1: Identify the system of inequalities.
The given constraints are: 1) $4x+3y\le60$, 2) $y\ge2x$, 3) $x\ge3$, and the non-negativity constraints $x\ge0$, $y\ge0$.

Step 2: Analyze the boundary conditions for the specific regions.
Looking at a standard graph for these equations, the region $S_{2}$ is typically bounded between the lines $x=3$, $y=2x$, and $4x+3y=60$. To verify, we select a test point that lies comfortably inside the expected region.

Step 3: Select and test a coordinate point.
Let's select the test point $(4, 10)$. We will substitute $x=4$ and $y=10$ into each inequality to see if they all hold true.

Step 4: Evaluate the first two inequalities.
Substitute into $4x+3y\le60$: $4(4) + 3(10) = 16 + 30 = 46$. Since $46 \le 60$ is true, the point satisfies the first condition. Substitute into $y\ge2x$: $10 \ge 2(4) \Rightarrow 10 \ge 8$. This is also true.

Step 5: Evaluate the remaining constraints to finalize the region.
Check $x\ge3$: Since $4 \ge 3$, this is true. Check non-negativity: $4 \ge 0$ and $10 \ge 0$ are both true. Because the test point satisfies all given inequalities simultaneously, the region containing this point is the correct solution set. $$ \therefore \text{The solution set is represented by } S_{2} \text{ region.} $$