1. **Stating the problem:** We need to find the area of the quadrilateral formed under the line from (0,0) to (50,5) and including the shaded right triangle between points (20,0), (20,2), and (40,4). 2. **Understanding the graph and shape:** The line is linear from (0,0) to (50,5), so its equation is found by slope formula: $$\text{slope} = \frac{5-0}{50-0} = \frac{5}{50} = \frac{1}{10}$$ Equation of line: $$y = \frac{1}{10}x$$ 3. **Area under the line from 0 to 50:** This forms a triangle with base 50 and height 5. $$\text{Area} = \frac{1}{2} \times 50 \times 5 = 125$$ 4. **Shaded right triangle area:** Points are (20,0), (20,2), and (40,4). Base is vertical segment from (20,0) to (20,2) length 2, height is horizontal segment from (20,2) to (40,4). Since line is $y=\frac{1}{10}x$, at $x=40$, $y=4$, so height is 20 units horizontally. Area of shaded triangle: $$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 20 = 20$$ 5. **Quadrilateral area:** The quadrilateral is the area under the line from $x=20$ to $x=40$ minus the shaded triangle area. Area under line from 20 to 40: $$\text{Area} = \int_{20}^{40} \frac{1}{10}x \, dx = \left[ \frac{1}{20}x^2 \right]_{20}^{40} = \frac{1}{20}(40^2 - 20^2) = \frac{1}{20}(1600 - 400) = \frac{1200}{20} = 60$$ 6. **Subtract shaded triangle area from this to get quadrilateral area:** $$60 - 20 = 40$$ **Final answer:** The area of the quadrilateral is $40$ (thousands of dollars per day times days).