If `x , y, z in R, x + y + z= 4 and x^(2) + y^(2) + z^(2) = 6`, then the maximum possible value of z is

Options

A. 1
B. 2
C. `(3)/(2)`
D. `(4)/(3)`

Answer

To solve the problem step by step, we need to find the maximum possible value of \( z \) given the equations:

1. \( x + y + z = 4 \)
2. \( x^2 + y^2 + z^2 = 6 \)

### Step 1: Express \( x + y \) in terms of \( z \)

From the first equation, we can express \( x + y \) as:
\[ x + y = 4 - z \]

### Step 2: Express \( x^2 + y^2 \) in terms of \( z \)

Using the identity \( x^2 + y^2 = (x + y)^2 - 2xy \), we can substitute \( x + y \):
\[ x^2 + y^2 = (4 - z)^2 - 2xy \]

### Step 3: Substitute into the second equation

Now, substitute \( x^2 + y^2 \) into the second equation:
\[ (4 - z)^2 - 2xy + z^2 = 6 \]

### Step 4: Simplify the equation

Expanding \( (4 - z)^2 \):
\[ 16 - 8z + z^2 - 2xy + z^2 = 6 \]
Combine like terms:
\[ 2z^2 - 8z + 16 - 2xy = 6 \]
Rearranging gives:
\[ 2z^2 - 8z + 10 - 2xy = 0 \]
Thus,
\[ 2xy = 2z^2 - 8z + 10 \]
or
\[ xy = z^2 - 4z + 5 \]

### Step 5: Use the discriminant condition

For \( x \) and \( y \) to be real numbers, the quadratic equation \( t^2 - (4 - z)t + (z^2 - 4z + 5) = 0 \) must have a non-negative discriminant:
\[ D = (4 - z)^2 - 4(z^2 - 4z + 5) \geq 0 \]

### Step 6: Calculate the discriminant

Calculating the discriminant:
\[ D = (4 - z)^2 - 4(z^2 - 4z + 5) \]
Expanding:
\[ D = (16 - 8z + z^2) - (4z^2 - 16z + 20) \]
Combine like terms:
\[ D = 16 - 8z + z^2 - 4z^2 + 16z - 20 \]
\[ D = -3z^2 + 8z - 4 \]

### Step 7: Set the discriminant greater than or equal to zero

Now, we need:
\[ -3z^2 + 8z - 4 \geq 0 \]
Multiplying through by -1 (and reversing the inequality):
\[ 3z^2 - 8z + 4 \leq 0 \]

### Step 8: Find the roots of the quadratic

Using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot 4}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{8 \pm \sqrt{64 - 48}}{6} = \frac{8 \pm \sqrt{16}}{6} = \frac{8 \pm 4}{6} \]
Thus, the roots are:
\[ z = \frac{12}{6} = 2 \quad \text{and} \quad z = \frac{4}{6} = \frac{2}{3} \]

### Step 9: Determine the intervals

The quadratic \( 3z^2 - 8z + 4 \) opens upwards, so it is less than or equal to zero between the roots:
\[ \frac{2}{3} \leq z \leq 2 \]

### Conclusion

The maximum possible value of \( z \) is:
\[ \boxed{2} \]