If `x+y +z= 21` then find the maximum value of `(x-2) (y-1) (z+ 9)`

Answer

To find the maximum value of the expression \((x-2)(y-1)(z+9)\) given that \(x + y + z = 21\), we can use the method of Lagrange multipliers or apply the AM-GM inequality. Here, we will use the AM-GM inequality for simplicity.

### Step-by-Step Solution:

1. **Rearranging the Variables**:
We can express the variables in terms of their shifts:
Let \(a = x - 2\), \(b = y - 1\), and \(c = z + 9\).
Then, we can rewrite the original equation:
\[ x + y + z = 21 \implies (a + 2) + (b + 1) + (c - 9) = 21 \]
Simplifying this gives:
\[ a + b + c - 6 = 21 \implies a + b + c = 27 \]

2. **Applying AM-GM Inequality**:
According to the AM-GM inequality:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
Substituting \(a + b + c = 27\):
\[ \frac{27}{3} \geq \sqrt[3]{abc} \implies 9 \geq \sqrt[3]{abc} \]
Cubing both sides gives:
\[ 729 \geq abc \]

3. **Finding the Maximum Value**:
The maximum value of \((x-2)(y-1)(z+9)\) is thus \(abc\), which can be at most \(729\).

4. **Equality Condition**:
The equality in the AM-GM inequality holds when \(a = b = c\). Therefore, we set:
\[ a = b = c = 9 \]
This gives:
\[ x - 2 = 9 \implies x = 11 \]
\[ y - 1 = 9 \implies y = 10 \]
\[ z + 9 = 9 \implies z = 0 \]

5. **Final Verification**:
We check:
\[ x + y + z = 11 + 10 + 0 = 21 \]
This satisfies the original condition.

Thus, the maximum value of \((x-2)(y-1)(z+9)\) is:
\[ \boxed{729} \]