If `a, b, c in R^(+)` such that `a+b+c=27`, then the maximum value of `a^(2)b^(3)c^(4)` is equal to

Options

A. `2^(8). 3^(10)`
B. `2^(9).3^(12)`
C. `2^(10).3^(12)`
D. `2^(11).3^(13)`

Answer

To find the maximum value of \( a^2 b^3 c^4 \) given that \( a + b + c = 27 \) and \( a, b, c \in \mathbb{R}^+ \), 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. **Express the constraint**: We know that \( a + b + c = 27 \).

2. **Rewrite the expression**: We want to maximize \( a^2 b^3 c^4 \). To apply the AM-GM inequality, we can express the terms in a way that matches their powers:
\[ a^2 b^3 c^4 = a^{2} \cdot b^{3} \cdot c^{4} \]

3. **Set up the AM-GM inequality**: We can rewrite the terms as follows:
\[ \frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} = 27 \]
This gives us 9 terms in total.

4. **Apply AM-GM**: By the AM-GM inequality, we have:
\[ \frac{\frac{a}{2} + \frac{a}{2} + \frac{b}{3} + \frac{b}{3} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4} + \frac{c}{4}}{9} \geq \sqrt[9]{\left(\frac{a}{2}\right)^2 \left(\frac{b}{3}\right)^2 \left(\frac{c}{4}\right)^4} \]

5. **Calculate the left-hand side**: Since \( a + b + c = 27 \), we have:
\[ \frac{27}{9} = 3 \]

6. **Set up the inequality**: Thus, we can write:
\[ 3 \geq \sqrt[9]{\left(\frac{a}{2}\right)^2 \left(\frac{b}{3}\right)^3 \left(\frac{c}{4}\right)^4} \]

7. **Raise both sides to the 9th power**: This gives:
\[ 3^9 \geq \left(\frac{a}{2}\right)^2 \left(\frac{b}{3}\right)^3 \left(\frac{c}{4}\right)^4 \]

8. **Express the right-hand side**: We can express this as:
\[ 3^9 \geq \frac{a^2 b^3 c^4}{2^2 \cdot 3^3 \cdot 4^4} \]

9. **Simplify the inequality**: Rearranging gives:
\[ a^2 b^3 c^4 \leq 3^9 \cdot 2^2 \cdot 3^3 \cdot 4^4 \]

10. **Calculate the maximum value**: Now, we need to compute \( 3^9 \cdot 2^2 \cdot 3^3 \cdot 4^4 \):
- \( 3^9 = 19683 \)
- \( 2^2 = 4 \)
- \( 3^3 = 27 \)
- \( 4^4 = 256 \)

Therefore,
\[ a^2 b^3 c^4 \leq 19683 \cdot 4 \cdot 27 \cdot 256 \]

11. **Final calculation**:
\[ 19683 \cdot 4 = 78732 \]
\[ 78732 \cdot 27 = 2125764 \]
\[ 2125764 \cdot 256 = 544864256 \]

Thus, the maximum value of \( a^2 b^3 c^4 \) is \( 544864256 \).