If `x+y=a`, then the greatest value of `x^(2)y^(3)` is

Options

A. `(108a^(5))/(3125)`
B. `(108a^(5))/(625)`
C. `(54a^(5))/(3125)`
D. `(108a^(4))/(3125)`

Explanation

To find the greatest value of \( x^2 y^3 \) given that \( x + y = a \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Express the variables**: We know that \( x + y = a \). We want to maximize \( x^2 y^3 \).

2. **Apply AM-GM Inequality**: We can apply the AM-GM inequality to the terms we have. To do this, we can rewrite \( x^2 \) as \( x + x \) (two times) and \( y^3 \) as \( y + y + y \) (three times). Thus, we can express our terms as:
\[ x + x + y + y + y \]

3. **Calculate the AM**: The Arithmetic Mean of these five terms is:
\[ \frac{x + x + y + y + y}{5} = \frac{2x + 3y}{5} \]

4. **Use the constraint**: We know that \( x + y = a \). We can express \( y \) in terms of \( x \):
\[ y = a - x \]
Substituting this into the AM gives:
\[ \frac{2x + 3(a - x)}{5} = \frac{2x + 3a - 3x}{5} = \frac{3a - x}{5} \]

5. **Apply AM-GM**: According to the AM-GM inequality:
\[ \frac{2x + 3y}{5} \geq \sqrt[5]{x^2 y^3} \]
Therefore, we have:
\[ \frac{3a - x}{5} \geq \sqrt[5]{x^2 y^3} \]

6. **Raise both sides to the power of 5**: This gives us:
\[ \left(\frac{3a - x}{5}\right)^5 \geq x^2 y^3 \]

7. **Find maximum value**: To find the maximum value of \( x^2 y^3 \), we need to find the optimal values of \( x \) and \( y \) that satisfy \( x + y = a \). From the AM-GM equality condition, we know that equality holds when all terms are equal:
\[ x = x = y = y = y \]
This implies:
\[ 2x = 3y \implies x = \frac{3}{2}y \]

8. **Substituting back**: Substitute \( y = a - x \) into \( x = \frac{3}{2}y \):
\[ x = \frac{3}{2}(a - x) \]
Solving for \( x \):
\[ 2x = 3a - 3x \implies 5x = 3a \implies x = \frac{3a}{5} \]
Then, substituting \( x \) back to find \( y \):
\[ y = a - x = a - \frac{3a}{5} = \frac{2a}{5} \]

9. **Calculate \( x^2 y^3 \)**: Now substituting \( x \) and \( y \) back into \( x^2 y^3 \):
\[ x^2 y^3 = \left(\frac{3a}{5}\right)^2 \left(\frac{2a}{5}\right)^3 = \frac{9a^2}{25} \cdot \frac{8a^3}{125} = \frac{72a^5}{3125} \]

10. **Final Result**: Therefore, the greatest value of \( x^2 y^3 \) is:
\[ \frac{72a^5}{3125} \]