Find minimum value of `px+qy` where `p>0, q>0, x>0, y>0` when `xy=r,^2` without using derivatives.

Options

A. `pqsqrtr`
B. `2pqsqrtr`
C. `2rsqrtpq`
D. None of these

Answer

To find the minimum value of \( px + qy \) given that \( xy = r^2 \) and \( p > 0, q > 0, x > 0, y > 0 \) without using derivatives, we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the AM-GM Inequality**:
The AM-GM inequality states that for any two positive numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
with equality when \( a = b \).

2. **Setting Up the Problem**:
Let \( a = px \) and \( b = qy \). Since \( p > 0 \), \( q > 0 \), \( x > 0 \), and \( y > 0 \), both \( a \) and \( b \) are positive.

3. **Applying AM-GM Inequality**:
According to the AM-GM inequality:
\[ \frac{px + qy}{2} \geq \sqrt{(px)(qy)} \]
This can be rewritten as:
\[ px + qy \geq 2\sqrt{pqxy} \]

4. **Substituting the Given Condition**:
We know from the problem that \( xy = r^2 \). Substituting this into the inequality gives:
\[ px + qy \geq 2\sqrt{pq(r^2)} \]

5. **Simplifying the Expression**:
Since \( \sqrt{r^2} = r \), we can simplify the inequality:
\[ px + qy \geq 2r\sqrt{pq} \]

6. **Conclusion**:
The minimum value of \( px + qy \) occurs when \( px = qy \), which is when \( a = b \) in the AM-GM inequality. Thus, the minimum value of \( px + qy \) is:
\[ \text{Minimum value} = 2r\sqrt{pq} \]

### Final Result:
The minimum value of \( px + qy \) is \( 2r\sqrt{pq} \).