Solution: The maximum occurs at the vertex $ x = -\fracb2a = -\frac182(-3) = 3 $. Substitute $ x = 3 $ into $ Y $: $ Y = -3(9) + 18(3) + 20 = -27 + 54 + 20 = 47 $. The fertilizer amount maximizing yield is $ \boxed3 $ kg. - Abu Waleed Tea

Understanding the Context

Assume crop yield $ Y $ is modeled by a quadratic equation:
$$
Y = -3x^2 + 18x + 20
$$
where $ x $ represents the amount of fertilizer applied in kilograms. This quadratic function reflects how yield typically rises to a peak before decreasing—due to diminishing returns on excessive fertilizer use.

Finding the Maximum Yield at the Vertex
The vertex of a parabola given by $ Y = ax^2 + bx + c $ occurs at:
$$
x = -rac{b}{2a}
$$
Substituting $ a = -3 $ and $ b = 18 $:
$$
x = -rac{18}{2(-3)} = -rac{18}{-6} = 3
$$
Thus, the maximum yield is achieved when farmers apply exactly 3 kg of fertilizer.

Calculating the Maximum Yield
To determine the actual yield at this optimal point, substitute $ x = 3 $ back into the yield equation:
$$
Y = -3(3)^2 + 18(3) + 20 = -3(9) + 54 + 20 = -27 + 54 + 20 = 47
$$
This means the maximum yield is 47 units (e.g., kilograms of produce), achieved using 3 kg of fertilizer.

Conclusion: The optimal fertilizer amount for peak yield is clearly 3 kg, scientifically proven by vertex analysis. Applying this precise dosage not only maximizes output but also prevents resource overuse, supporting sustainable and profitable farming practices.

Key Insights

For consistent results, always measure and adjust fertilizer according to this ideal point—ensuring your investment delivers maximum return.

Boxed Final Answer:
The fertilizer amount maximizing yield is $ oxed{3} $ kg.