If `x,y,z` be three positive numbers such that `xyz^(2)` has the greatest value `(1)/(64)`, then the value of `(1)/(x)+(1)/(y)+(1)/(z)` is

Options

A. `6`
B. `8`
C. `10`
D. `12`

Explanation

To solve the problem, we need to find the value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) given that \( xyz^2 \) has a maximum value of \( \frac{1}{64} \).

### Step-by-Step Solution:

1. **Understand the Given Condition**:
We know that \( xyz^2 \leq \frac{1}{64} \). To maximize \( xyz^2 \), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

2. **Apply AM-GM Inequality**:
According to the AM-GM inequality, for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. We can express \( z^2 \) as \( z \cdot z \):
\[ \frac{x + y + z + z}{4} \geq \sqrt[4]{xyz^2} \]
This implies:
\[ \frac{x + y + 2z}{4} \geq \sqrt[4]{xyz^2} \]

3. **Substituting the Maximum Value**:
Since \( xyz^2 \) has a maximum value of \( \frac{1}{64} \):
\[ \sqrt[4]{xyz^2} \leq \sqrt[4]{\frac{1}{64}} = \frac{1}{4} \]
Therefore, we have:
\[ \frac{x + y + 2z}{4} \geq \frac{1}{4} \]
Multiplying both sides by 4 gives:
\[ x + y + 2z \geq 1 \]

4. **Finding the Maximum Product**:
The product \( xyz^2 \) is maximized when \( x = y = z \). Let \( z = 2k \) and \( x = y = k \) for some \( k \). Then:
\[ xyz^2 = k \cdot k \cdot (2k)^2 = 4k^4 \]
Setting this equal to \( \frac{1}{64} \):
\[ 4k^4 = \frac{1}{64} \implies k^4 = \frac{1}{256} \implies k = \frac{1}{4} \]

5. **Calculating Values of x, y, z**:
Thus, we have:
\[ x = \frac{1}{4}, \quad y = \frac{1}{4}, \quad z = 2k = \frac{1}{2} \]

6. **Finding \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \)**:
Now we can calculate:
\[ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{\frac{1}{4}} + \frac{1}{\frac{1}{4}} + \frac{1}{\frac{1}{2}} = 4 + 4 + 2 = 10 \]

### Final Answer:
The value of \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \) is \( 10 \).

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