If `a gt 0,bgt 0,cgt0 and 2a +b+3c=1`, then

Options

A. `a^4b^2c^2` is greatest then `a=(1)/(4)`
B. `a^4b^2c^2` is greatest then `b=(1)/(4)`
C. `a^4b^2c^2` is greatest then `c=(1)/(12)`
D. greatest value of `a^4b^2c^2 is (1)/(9.4^8)`

Explanation

To solve the problem given \( 2a + b + 3c = 1 \) with the constraints \( a > 0, b > 0, c > 0 \), we will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Rewrite the equation**: We start with the equation \( 2a + b + 3c = 1 \).

2. **Apply AM-GM Inequality**: To apply the AM-GM inequality, we can split the terms appropriately. We can express \( 2a \) as \( a + a \), and \( 3c \) as \( c + c + c \). Thus, we rewrite the equation as:
\[ a + a + b + c + c + c = 1 \]
This gives us a total of 6 terms.

3. **Using AM-GM**: According to the AM-GM inequality, the arithmetic mean of non-negative numbers is greater than or equal to their geometric mean. Therefore, we have:
\[ \frac{a + a + b + c + c + c}{6} \geq \sqrt[6]{a^2 \cdot b \cdot c^3} \]
Substituting the left-hand side:
\[ \frac{1}{6} \geq \sqrt[6]{a^2 \cdot b \cdot c^3} \]

4. **Raise both sides to the power of 6**: To eliminate the root, we raise both sides to the power of 6:
\[ \left(\frac{1}{6}\right)^6 \geq a^2 \cdot b \cdot c^3 \]
Simplifying the left side:
\[ \frac{1}{6^6} \geq a^2 \cdot b \cdot c^3 \]

5. **Expressing the result**: The inequality we derived gives us a relationship between \( a, b, \) and \( c \). We can express the maximum value of \( a^2 b c^3 \):
\[ a^2 b c^3 \leq \frac{1}{6^6} \]

6. **Finding the maximum value**: To find the maximum value of \( a^4 b^2 c^2 \), we can use the relationship we derived:
\[ a^4 b^2 c^2 = (a^2 b c^3) \cdot \frac{c^2}{c} \]
Since \( c \) is positive, we can maximize \( a^4 b^2 c^2 \) using the values of \( a, b, c \) derived from the equality case of AM-GM.

7. **Equality case**: The equality in AM-GM holds when all terms are equal:
\[ a = a = b = c = k \]
Substituting into the equation gives:
\[ 2k + k + 3k = 1 \implies 6k = 1 \implies k = \frac{1}{6} \]
Thus, \( a = \frac{1}{6}, b = \frac{1}{6}, c = \frac{1}{6} \).

8. **Calculating maximum value**: Substitute \( k \) back into \( a^4 b^2 c^2 \):
\[ a^4 b^2 c^2 = \left(\frac{1}{6}\right)^4 \left(\frac{1}{6}\right)^2 \left(\frac{1}{6}\right)^2 = \frac{1}{6^8} \]

### Conclusion:
The maximum value of \( a^4 b^2 c^2 \) given the constraints is \( \frac{1}{6^8} \).