Daerah penyelesaian dari sistem pertidaksamaan: 7x + 2y ≤ 14; 3x + 5y ≥ 15; x ≥ 0, y ≥ 0. Pada gambar di bawah ini adalah ....

The system of inequalities is:

1. $7x + 2y \le 14$
2. $3x + 5y \ge 15$
3. $x \ge 0$
4. $y \ge 0$

For the first inequality, $7x + 2y = 14$:

• If $x = 0$, then $2y = 14 \Rightarrow y = 7$. So, the line passes through $(0, 7)$.
• If $y = 0$, then $7x = 14 \Rightarrow x = 2$. So, the line passes through $(2, 0)$. This is the steeper line on the graph, connecting $(0,7)$ and $(2,0)$. Let's call this $L_1$.

For the second inequality, $3x + 5y = 15$:

• If $x = 0$, then $5y = 15 \Rightarrow y = 3$. So, the line passes through $(0, 3)$.
• If $y = 0$, then $3x = 15 \Rightarrow x = 5$. So, the line passes through $(5, 0)$. This is the less steep line on the graph, connecting $(0,3)$ and $(5,0)$. Let's call this $L_2$.

• For $7x + 2y \le 14$: Test the origin $(0,0)$. $7(0) + 2(0) = 0 \le 14$. This is true. So, the solution region for this inequality is the area below or to the left of $L_1$.
• For $3x + 5y \ge 15$: Test the origin $(0,0)$. $3(0) + 5(0) = 0 \ge 15$. This is false. So, the solution region for this inequality is the area above or to the right of $L_2$.
• For $x \ge 0$: This means the region is to the right of the y-axis.
• For $y \ge 0$: This means the region is above the x-axis.

We are looking for the region that satisfies all four conditions:

1. Below or to the left of $L_1$ (the line through $(0,7)$ and $(2,0)$).
2. Above or to the right of $L_2$ (the line through $(0,3)$ and $(5,0)$).
3. To the right of the y-axis ($x \ge 0$).
4. Above the x-axis ($y \ge 0$).

Let's examine the labeled regions on the graph:

• Region I: This region is below $L_2$. It does not satisfy $3x + 5y \ge 15$.
• Region II: This region is below $L_2$. It does not satisfy $3x + 5y \ge 15$.
• Region III: This region is above $L_1$. It does not satisfy $7x + 2y \le 14$.
• Region IV: This region is above $L_1$. It does not satisfy $7x + 2y \le 14$.
• Region V: This region is bounded by the y-axis, $L_1$, and $L_2$. Let's test a point in this region, for example, $(0.5, 4)$.
  • $7(0.5) + 2(4) = 3.5 + 8 = 11.5$. Since $11.5 \le 14$, the first inequality is satisfied.
  • $3(0.5) + 5(4) = 1.5 + 20 = 21.5$. Since $21.5 \ge 15$, the second inequality is satisfied.
  • $x = 0.5 \ge 0$ and $y = 4 \ge 0$, so the third and fourth inequalities are satisfied. Since all conditions are met, Region V is the solution area.

The final answer is $\text{E. V}$.