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

Options

A. `2r sqrt(pq)`
B. `2pqsqrtr`
C. `-2rsqrt(pq)`
D. none of these

Answer

To find the minimum value of \( px + qy \) under the constraint \( xy = r^2 \) where \( p > 0, q > 0, x > 0, y > 0 \), we can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to minimize the expression \( px + qy \) given that \( xy = r^2 \).

2. **Using AM-GM Inequality**: According to the AM-GM inequality, for any non-negative numbers \( a \) and \( b \):
\[ \frac{a + b}{2} \geq \sqrt{ab} \]
with equality when \( a = b \).

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

4. **Applying AM-GM**: We apply the AM-GM inequality to \( a \) and \( b \):
\[ \frac{px + qy}{2} \geq \sqrt{(px)(qy)} \]
This simplifies to:
\[ px + qy \geq 2\sqrt{pqxy} \]

5. **Substituting the Constraint**: We know from the problem that \( xy = r^2 \). Substitute this into the inequality:
\[ px + qy \geq 2\sqrt{pq \cdot r^2} \]
This further simplifies to:
\[ px + qy \geq 2r\sqrt{pq} \]

6. **Conclusion**: The minimum value of \( px + qy \) occurs when \( px = qy \), which happens when \( x \) and \( y \) are chosen such that \( \frac{px}{qy} = 1 \) or \( px = qy \). Thus, the minimum value of \( px + qy \) is:
\[ \boxed{2r\sqrt{pq}} \]