If x,y are positive real numbers and `3x+4y=5`, then the lagest possible value of `16x^(2)y^(3)` is

Answer

To find the largest possible value of \( 16x^2y^3 \) given the constraint \( 3x + 4y = 5 \), we can use the method of Lagrange multipliers or the AM-GM inequality. Here, we will use the AM-GM inequality for simplicity.

### Step-by-step Solution:

1. **Understand the constraint**: We have the equation \( 3x + 4y = 5 \). This can be rearranged to express \( y \) in terms of \( x \):
\[ y = \frac{5 - 3x}{4} \]

2. **Express the function**: We need to maximize \( 16x^2y^3 \). Substituting the expression for \( y \) into this function gives:
\[ 16x^2y^3 = 16x^2 \left(\frac{5 - 3x}{4}\right)^3 \]

3. **Simplify the expression**: Expanding this, we have:
\[ 16x^2 \left(\frac{(5 - 3x)^3}{64}\right) = \frac{16x^2(5 - 3x)^3}{64} = \frac{x^2(5 - 3x)^3}{4} \]

4. **Apply AM-GM Inequality**: We can apply the AM-GM inequality to the terms \( \frac{3x}{2}, \frac{3x}{2}, \frac{4y}{3}, \frac{4y}{3}, \frac{4y}{3} \):
\[ \frac{\frac{3x}{2} + \frac{3x}{2} + \frac{4y}{3} + \frac{4y}{3} + \frac{4y}{3}}{5} \geq \sqrt[5]{\left(\frac{3x}{2}\right)^2 \left(\frac{4y}{3}\right)^3} \]

5. **Substitute the constraint**: The left-hand side simplifies to:
\[ \frac{5}{5} = 1 \]
Therefore, we have:
\[ 1 \geq \sqrt[5]{\left(\frac{3x}{2}\right)^2 \left(\frac{4y}{3}\right)^3} \]

6. **Raise both sides to the power of 5**: This leads to:
\[ 1 \geq \frac{(3x)^2(4y)^3}{2^2 \cdot 3^3} \]

7. **Rearranging gives**:
\[ 16x^2y^3 \leq \frac{1 \cdot 2^2 \cdot 3^3}{1} = \frac{4 \cdot 27}{1} = 108 \]

8. **Finding the maximum**: The maximum value occurs when equality holds in the AM-GM inequality, which happens when:
\[ \frac{3x}{2} = \frac{4y}{3} \]
Solving this along with the constraint \( 3x + 4y = 5 \) gives us the values of \( x \) and \( y \).

9. **Final calculation**: After solving, we find that the maximum value of \( 16x^2y^3 \) is \( 3 \).

### Conclusion:
The largest possible value of \( 16x^2y^3 \) given the constraint \( 3x + 4y = 5 \) is \( \boxed{3} \).