1. **Problem statement:** A gardener plants peas in 30 rows, each producing 4000 g. For every additional row, production per row decreases by 100 g. We need to model this with a quadratic function. 2. **Define variables:** Let $x$ be the number of additional rows beyond 30. 3. **Rows and production per row:** Total rows = $30 + x$. Production per row = $4000 - 100x$ grams. 4. **Total production function:** Total production $P(x) = (30 + x)(4000 - 100x)$ grams. 5. **Expand the function:** $$P(x) = 30 \times 4000 - 30 \times 100x + x \times 4000 - 100x^2 = 120000 - 3000x + 4000x - 100x^2$$ 6. **Simplify:** $$P(x) = -100x^2 + 1000x + 120000$$ 7. **Convert to kilograms:** Since 1000 g = 1 kg, $$P(x) = \frac{-100x^2 + 1000x + 120000}{1000} = -0.1x^2 + x + 120$$ kilograms. 8. **Maximum production:** The quadratic opens downward (coefficient of $x^2$ is negative), so maximum at vertex. 9. **Vertex formula:** $$x = -\frac{b}{2a} = -\frac{1}{2 \times (-0.1)} = \frac{1}{0.2} = 5$$ 10. **Maximum production value:** $$P(5) = -0.1(5)^2 + 5 + 120 = -0.1 \times 25 + 5 + 120 = -2.5 + 5 + 120 = 122.5$$ kilograms. 11. **Number of rows for maximum:** $$30 + 5 = 35$$ rows. 12. **Assumptions:** - The decrease in production per row is linear and constant. - The relationship between rows and production per row is accurately modeled by the given linear decrease. - No other factors affect production. Final answers: - a) Quadratic function: $$P(x) = -0.1x^2 + x + 120$$ kilograms, where $x$ is additional rows beyond 30. - b) Maximum production is 122.5 kilograms at 35 rows. - c) Assumptions as stated above.