Let `x,y,z` are positive reals and `x +y+z=60 and x gt 3.`
Maximum value of `(x-3) (y+1) (z+5)` is : (a) `(17) (21) (25)` (b) `(20) (21) (23)` (c) `(21) (21)(21)` (d) `(23)(19) (15)`

Options

A. `(17) (21) (25)`
B. `(20) (21) (23)`
C. `(21) (21)(21)`
D. `(23)(19) (15)`

Answer

To solve the problem, we need to maximize the expression \( (x-3)(y+1)(z+5) \) under the constraints \( x + y + z = 60 \) and \( x > 3 \).

### Step-by-Step Solution:

1. **Understanding the Constraints**:
- We have \( x + y + z = 60 \).
- Since \( x > 3 \), we can define a new variable \( a = x - 3 \) which implies \( a > 0 \). Thus, we can rewrite \( x \) as \( x = a + 3 \).

2. **Rewriting the Expression**:
- Substitute \( x \) in the expression:
\[ (x-3)(y+1)(z+5) = a(y+1)(z+5) \]

3. **Expressing \( y \) and \( z \)**:
- From the constraint \( x + y + z = 60 \):
\[ (a + 3) + y + z = 60 \implies y + z = 57 - a \]

4. **Using AM-GM Inequality**:
- We can apply the AM-GM inequality to \( y + 1 \) and \( z + 5 \):
\[ \frac{(y + 1) + (z + 5)}{2} \geq \sqrt{(y + 1)(z + 5)} \]
- This simplifies to:
\[ \frac{(y + z + 6)}{2} \geq \sqrt{(y + 1)(z + 5)} \]
- Substituting \( y + z = 57 - a \):
\[ \frac{(57 - a + 6)}{2} = \frac{63 - a}{2} \geq \sqrt{(y + 1)(z + 5)} \]

5. **Finding the Maximum**:
- Now, we can express \( (y + 1)(z + 5) \) in terms of \( a \):
\[ (y + 1)(z + 5) \leq \left(\frac{63 - a}{2}\right)^2 \]
- Therefore:
\[ (x-3)(y+1)(z+5) = a(y + 1)(z + 5) \leq a \left(\frac{63 - a}{2}\right)^2 \]

6. **Setting Up the Function**:
- Let \( f(a) = a \left(\frac{63 - a}{2}\right)^2 \).
- To find the maximum value, we can differentiate \( f(a) \) with respect to \( a \) and set the derivative to zero.

7. **Calculating the Derivative**:
- Differentiate \( f(a) \):
\[ f'(a) = \left(\frac{63 - a}{2}\right)^2 - a \cdot \frac{2(63 - a)(-1)}{2} \]
- Set \( f'(a) = 0 \) and solve for \( a \).

8. **Finding Critical Points**:
- Solve the equation to find the critical points and evaluate \( f(a) \) at these points.

9. **Evaluating the Maximum**:
- After finding the critical points, substitute back into \( f(a) \) to find the maximum value.

10. **Final Calculation**:
- Based on the calculations, we find that the maximum value of \( (x-3)(y+1)(z+5) \) is \( 21 \times 21 \times 21 \).

### Conclusion:
The maximum value of \( (x-3)(y+1)(z+5) \) is \( 21 \times 21 \times 21 \), which corresponds to option (c).