Graph the feasible region for the following system of inequalities: 0.25x + 0.60y <= 75, 0.60x + 0.20y <= 90, x >= 0, y >= 0.

Draw a coordinate plane with the x-axis representing the number of barrels of pesticide A ($x$) and the y-axis representing the number of barrels of pesticide B ($y$). Since $x \ge 0$ and $y \ge 0$, focus on the first quadrant.

• To find the x-intercept, set $y=0$: $0.25x = 75 \implies x = \frac{75}{0.25} = 300$. Plot $(300, 0)$.
• To find the y-intercept, set $x=0$: $0.60y = 75 \implies y = \frac{75}{0.60} = 125$. Plot $(0, 125)$.
• Draw a solid line connecting $(300, 0)$ and $(0, 125)$.

• To find the x-intercept, set $y=0$: $0.60x = 90 \implies x = \frac{90}{0.60} = 150$. Plot $(150, 0)$.
• To find the y-intercept, set $x=0$: $0.20y = 90 \implies y = \frac{90}{0.20} = 450$. Plot $(0, 450)$.
• Draw a solid line connecting $(150, 0)$ and $(0, 450)$.

Solve the system of equations:

1. $0.25x + 0.60y = 75$
2. $0.60x + 0.20y = 90$

From previous calculations, the intersection point is approximately $(125.81, 72.58)$. Plot this point.

• For $0.25x + 0.60y \le 75$, shade the region below the line connecting $(300,0)$ and $(0,125)$.
• For $0.60x + 0.20y \le 90$, shade the region below the line connecting $(150,0)$ and $(0,450)$.
• Since $x \ge 0$ and $y \ge 0$, the feasible region is restricted to the first quadrant.

The feasible region is the area in the first quadrant that is below both lines. It is a polygon with the following vertices:

• $(0,0)$
• $(150,0)$
• $\left(\frac{3900}{31}, \frac{2250}{31}\right) \approx (125.81, 72.58)$
• $(0,125)$

This shaded region represents all possible combinations of pesticide A and B production that satisfy the pollution laws.