If a+b+c=13 What is the maximum value of (a-3)(b-2)(c+1)?

Options

A. 26
B. 27
C. 30
D. 19

Answer

To find the maximum value of the expression \((a-3)(b-2)(c+1)\) given that \(a + b + c = 13\), we can use the method of Lagrange multipliers or apply the AM-GM inequality. Here, we will use the AM-GM inequality approach.

### Step-by-Step Solution:

1. **Rewrite the Expression**:
We want to maximize \((a-3)(b-2)(c+1)\). Let's denote:
\[ x = a - 3, \quad y = b - 2, \quad z = c + 1 \]
Then we can express \(a\), \(b\), and \(c\) in terms of \(x\), \(y\), and \(z\):
\[ a = x + 3, \quad b = y + 2, \quad c = z - 1 \]
Substituting these into the constraint \(a + b + c = 13\):
\[ (x + 3) + (y + 2) + (z - 1) = 13 \]
Simplifying this gives:
\[ x + y + z + 4 = 13 \implies x + y + z = 9 \]

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. Thus:
\[ \frac{x + y + z}{3} \geq \sqrt[3]{xyz} \]
Substituting \(x + y + z = 9\):
\[ \frac{9}{3} \geq \sqrt[3]{xyz} \implies 3 \geq \sqrt[3]{xyz} \]
Cubing both sides gives:
\[ 27 \geq xyz \]

3. **Conclusion**:
The maximum value of \((a-3)(b-2)(c+1)\) is therefore:
\[ (a-3)(b-2)(c+1) \leq 27 \]
The maximum value occurs when \(x = y = z\). Thus, we can set:
\[ x = y = z = 3 \quad \text{(since } x + y + z = 9\text{)} \]
This implies:
\[ a - 3 = 3 \implies a = 6, \quad b - 2 = 3 \implies b = 5, \quad c + 1 = 3 \implies c = 2 \]
Therefore, the maximum value of \((a-3)(b-2)(c+1)\) is:
\[ \boxed{27} \]