If `p^(2)+q^(2)+r^(2)=1`, then the maximum vlaue of pqr is :

Options

A. 1
B. `(1)/(3sqrt3)`
C. 3
D. none of these

Answer

To find the maximum value of \( pqr \) given the constraint \( p^2 + q^2 + r^2 = 1 \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Here’s a step-by-step solution:

### Step 1: Apply the AM-GM Inequality
We start with the AM-GM inequality for three non-negative numbers \( p^2, q^2, r^2 \):
\[ \frac{p^2 + q^2 + r^2}{3} \geq \sqrt[3]{p^2 q^2 r^2} \]
Given that \( p^2 + q^2 + r^2 = 1 \), we can substitute this into the inequality:
\[ \frac{1}{3} \geq \sqrt[3]{p^2 q^2 r^2} \]

### Step 2: Raise Both Sides to the Power of 3
To eliminate the cube root, we raise both sides to the power of 3:
\[ \left(\frac{1}{3}\right)^3 \geq p^2 q^2 r^2 \]
This simplifies to:
\[ \frac{1}{27} \geq p^2 q^2 r^2 \]

### Step 3: Relate \( pqr \) to \( p^2 q^2 r^2 \)
Notice that:
\[ p^2 q^2 r^2 = (pqr)^2 \]
Thus, we can write:
\[ \frac{1}{27} \geq (pqr)^2 \]

### Step 4: Take the Square Root
Taking the square root of both sides gives:
\[ \sqrt{\frac{1}{27}} \geq pqr \]
This simplifies to:
\[ \frac{1}{3\sqrt{3}} \geq pqr \]

### Step 5: Conclusion
The maximum value of \( pqr \) is \( \frac{1}{3\sqrt{3}} \).

### Final Answer
Thus, the maximum value of \( pqr \) is:
\[ \boxed{\frac{1}{3\sqrt{3}}} \]

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