Let `x,y,z` are positive reals and `x +y+z=60 and x gt 3.`
Maximum value of xyz is :

Options

A. `8xx10^(3)`
B. `27xx10^(3)`
C. `64xx10^(3)`
D. `125xx10^(3)`

Answer

To find the maximum value of \( xyz \) given the constraints \( x + y + z = 60 \) and \( x > 3 \), we can use the method of inequalities, specifically the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Apply the AM-GM Inequality**:
The AM-GM inequality states that for any non-negative real numbers \( a_1, a_2, \ldots, a_n \):
\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \]
In our case, let \( a_1 = x \), \( a_2 = y \), and \( a_3 = z \). Thus, we have:
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]

2. **Substitute the Given Sum**:
We know that \( x + y + z = 60 \). Substituting this into the inequality gives:
\[ \frac{60}{3} \geq \sqrt[3]{xyz} \]
Simplifying this, we find:
\[ 20 \geq \sqrt[3]{xyz} \]

3. **Cubing Both Sides**:
To eliminate the cube root, we cube both sides:
\[ 20^3 \geq xyz \]
Calculating \( 20^3 \):
\[ 20^3 = 8000 \]
Thus, we have:
\[ xyz \leq 8000 \]

4. **Finding Conditions for Equality**:
The equality in the AM-GM inequality holds when all the numbers are equal, i.e., \( x = y = z \). Therefore, we set:
\[ x = y = z = k \]
From the constraint \( x + y + z = 60 \):
\[ 3k = 60 \implies k = 20 \]
Thus, \( x = y = z = 20 \) gives \( xyz = 20 \times 20 \times 20 = 8000 \).

5. **Check the Condition \( x > 3 \)**:
Since \( x = 20 \), it satisfies the condition \( x > 3 \).

### Conclusion:
The maximum value of \( xyz \) under the given constraints is \( 8000 \).