If x+y+z = 21, the maximum value of (x-6)(y+7)(z-4) is :

Options

A. 343
B. 216
C. 125
D. not unique

Answer

To find the maximum value of the expression \((x-6)(y+7)(z-4)\) given that \(x + y + z = 21\), we can follow these steps:

### Step 1: Rewrite the expression
We want to maximize the expression \((x-6)(y+7)(z-4)\). Let's denote:
- \(a = x - 6\)
- \(b = y + 7\)
- \(c = z - 4\)

Then, we can express \(x\), \(y\), and \(z\) in terms of \(a\), \(b\), and \(c\):
- \(x = a + 6\)
- \(y = b - 7\)
- \(z = c + 4\)

### Step 2: Substitute into the constraint
Substituting these into the constraint \(x + y + z = 21\):
\[ (a + 6) + (b - 7) + (c + 4) = 21 \]
Simplifying this gives:
\[ a + b + c + 3 = 21 \]
Thus,
\[ a + b + c = 18 \]

### Step 3: Apply the method of Lagrange multipliers or AM-GM inequality
To maximize the product \(abc\) under the constraint \(a + b + c = 18\), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality, which states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean.

By AM-GM:
\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]
Substituting \(a + b + c = 18\):
\[ \frac{18}{3} \geq \sqrt[3]{abc} \]
This simplifies to:
\[ 6 \geq \sqrt[3]{abc} \]
Cubing both sides gives:
\[ 216 \geq abc \]

### Step 4: Find when equality holds
Equality in AM-GM holds when \(a = b = c\). Therefore, we set:
\[ a = b = c = 6 \]

### Step 5: Substitute back to find \(x\), \(y\), and \(z\)
Now substituting back:
- \(a = 6 \Rightarrow x - 6 = 6 \Rightarrow x = 12\)
- \(b = 6 \Rightarrow y + 7 = 6 \Rightarrow y = -1\)
- \(c = 6 \Rightarrow z - 4 = 6 \Rightarrow z = 10\)

### Step 6: Calculate the maximum value
Now we can calculate the maximum value of the expression:
\[ (x-6)(y+7)(z-4) = (12-6)(-1+7)(10-4) = 6 \cdot 6 \cdot 6 = 216 \]

### Conclusion
Thus, the maximum value of \((x-6)(y+7)(z-4)\) given \(x + y + z = 21\) is **216**.

---