Let `(p, q, r)` be a point on the plane `2x+2y+z=6`, then the least value of `p^2+q^2+r^2` is equal to

Options

A. `4`
B. `5`
C. `6`
D. `8`

Answer

To find the least value of \( p^2 + q^2 + r^2 \) for a point \( (p, q, r) \) on the plane defined by the equation \( 2x + 2y + z = 6 \), we can follow these steps:

### Step 1: Rewrite the Plane Equation
The given plane equation can be rewritten in a more manageable form:
\[ 2x + 2y + z = 6 \]
This can be rearranged to express \( z \):
\[ z = 6 - 2x - 2y \]

### Step 2: Set Up the Function to Minimize
We need to minimize the expression \( p^2 + q^2 + r^2 \). Substituting \( r \) with \( 6 - 2p - 2q \):
\[ p^2 + q^2 + r^2 = p^2 + q^2 + (6 - 2p - 2q)^2 \]

### Step 3: Expand the Expression
Now we expand the expression:
\[ = p^2 + q^2 + (6 - 2p - 2q)^2 \]
\[ = p^2 + q^2 + (36 - 24p - 24q + 4p^2 + 8pq + 4q^2) \]
\[ = 5p^2 + 5q^2 + 8pq - 24p - 24q + 36 \]

### Step 4: Use Partial Derivatives to Find Critical Points
To find the minimum, we take the partial derivatives with respect to \( p \) and \( q \) and set them to zero.

1. **Partial Derivative with respect to \( p \)**:
\[ \frac{\partial}{\partial p}(5p^2 + 5q^2 + 8pq - 24p - 24q + 36) = 10p + 8q - 24 = 0 \]

2. **Partial Derivative with respect to \( q \)**:
\[ \frac{\partial}{\partial q}(5p^2 + 5q^2 + 8pq - 24p - 24q + 36) = 10q + 8p - 24 = 0 \]

### Step 5: Solve the System of Equations
We have the following system of equations:
1. \( 10p + 8q = 24 \)
2. \( 10q + 8p = 24 \)

From the first equation:
\[ 10p + 8q = 24 \implies 5p + 4q = 12 \implies q = \frac{12 - 5p}{4} \]

Substituting \( q \) in the second equation:
\[ 10\left(\frac{12 - 5p}{4}\right) + 8p = 24 \]
\[ 30 - 12.5p + 8p = 24 \]
\[ -4.5p = -6 \implies p = \frac{4}{3} \]

Substituting \( p \) back to find \( q \):
\[ q = \frac{12 - 5(\frac{4}{3})}{4} = \frac{12 - \frac{20}{3}}{4} = \frac{\frac{36 - 20}{3}}{4} = \frac{16/3}{4} = \frac{4}{3} \]

### Step 6: Find \( r \)
Now substituting \( p \) and \( q \) into the equation for \( r \):
\[ r = 6 - 2p - 2q = 6 - 2(\frac{4}{3}) - 2(\frac{4}{3}) = 6 - \frac{16}{3} = \frac{18 - 16}{3} = \frac{2}{3} \]

### Step 7: Calculate \( p^2 + q^2 + r^2 \)
Now we calculate:
\[ p^2 + q^2 + r^2 = \left(\frac{4}{3}\right)^2 + \left(\frac{4}{3}\right)^2 + \left(\frac{2}{3}\right)^2 \]
\[ = \frac{16}{9} + \frac{16}{9} + \frac{4}{9} = \frac{36}{9} = 4 \]

### Conclusion
Thus, the least value of \( p^2 + q^2 + r^2 \) is:
\[ \boxed{4} \]