If x + y + z = 12 , then find the maximum value of (x – 1) (y – 2) (z – 3).

To find the maximum value of the product \((x-1)(y-2)(z-3)\) subject to the constraint \(x + y + z = 12\), we can use the method of Lagrange multipliers or directly apply the AM-GM inequality.

Let’s use the AM-GM inequality:

By the Arithmetic Mean-Geometric Mean inequality, for non-negative numbers \(a\), \(b\), and \(c\):

\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]

Equality holds when \(a = b = c\).

Applying this to our problem, we first rewrite \((x-1)(y-2)(z-3)\) in a form that allows us to use the AM-GM inequality.

Let \(a = x – 1\), \(b = y – 2\), and \(c = z – 3\). Then \(a + b + c = x + y + z – 6 = 12 – 6 = 6\).

By AM-GM, we have:

\[ \frac{a + b + c}{3} = \frac{6}{3} = 2 \geq \sqrt[3]{abc} \]

\[ \sqrt[3]{abc} \leq 2 \]

\[ abc \leq 8 \]

So the maximum value of \((x-1)(y-2)(z-3) = abc\) is \(8\), and this maximum is achieved when \(a = b = c = 2\), or equivalently, when \(x = 3\), \(y = 4\), and \(z = 5\).