Find maximum value of `a^(2)b^(3)` if `a+b=2`. When a & b are positive numbers

Options

A. `(2^(5)3^(5))/(2^(5))`
B. `(2^(7)3^(3))/(5^(5))`
C. `(2^(8)3^(2))/(5^(5))`
D. `(2^(6)3^(4))/(5^(5))`

Answer

To find the maximum value of \( a^2 b^3 \) given that \( a + b = 2 \) and both \( a \) and \( b \) are positive numbers, we can use the method of Lagrange multipliers or apply the AM-GM inequality. Here, we will use the AM-GM inequality.

### Step-by-Step Solution:

1. **Set Up the Problem**: We need to maximize \( a^2 b^3 \) under the constraint \( a + b = 2 \).

2. **Apply the AM-GM Inequality**: We can express \( a^2 b^3 \) in terms of the arithmetic mean and geometric mean. We can rewrite \( a^2 b^3 \) as follows:
\[ a^2 b^3 = \frac{a + a + b + b + b}{5} \cdot \frac{a + a + b + b + b}{5} \cdot \frac{a + a + b + b + b}{5} \]

3. **Calculate the Arithmetic Mean**: The arithmetic mean of \( a, a, b, b, b \) is:
\[ \frac{a + a + b + b + b}{5} = \frac{2 + 3b}{5} \]
Since \( a + b = 2 \), we can express \( b \) as \( b = 2 - a \).

4. **Substituting for \( b \)**: Substitute \( b \) into the arithmetic mean:
\[ \frac{2 + 3(2 - a)}{5} = \frac{2 + 6 - 3a}{5} = \frac{8 - 3a}{5} \]

5. **Using AM-GM**: According to the AM-GM inequality:
\[ \frac{a + a + b + b + b}{5} \geq \sqrt[5]{a^2 b^3} \]
Therefore,
\[ \frac{2 + 3b}{5} \geq \sqrt[5]{a^2 b^3} \]

6. **Finding Maximum Value**: To find the maximum value of \( a^2 b^3 \), we need to maximize \( \left( \frac{2 + 3b}{5} \right)^5 \). We know that the maximum occurs when all terms are equal, i.e., \( a = b \).

7. **Setting \( a = b \)**: If \( a = b \), then \( a + b = 2 \) implies \( 2a = 2 \) or \( a = 1 \) and \( b = 1 \).

8. **Calculating Maximum Value**: Substitute \( a = 1 \) and \( b = 1 \) into \( a^2 b^3 \):
\[ a^2 b^3 = 1^2 \cdot 1^3 = 1 \]

9. **Conclusion**: The maximum value of \( a^2 b^3 \) given the constraint \( a + b = 2 \) is:
\[ \boxed{1} \]