The minimum value of `px+py` when `xy=r^(2)` is equal to

Options

A. `2rsqrt(pq)`
B. `2pr sqrtr`
C. `-2rsqrt(pq)`
D. none of these

Answer

To find the minimum value of \( px + py \) given the constraint \( xy = r^2 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
From the constraint \( xy = r^2 \), we can express \( y \) as:
\[ y = \frac{r^2}{x} \]

### Step 2: Substitute \( y \) into the expression
Now substitute \( y \) into the expression \( px + py \):
\[ px + py = px + p\left(\frac{r^2}{x}\right) = px + \frac{pr^2}{x} \]
Thus, we can rewrite the expression as:
\[ f(x) = px + \frac{pr^2}{x} \]

### Step 3: Differentiate the function
To find the minimum value, we need to differentiate \( f(x) \):
\[ f'(x) = p - \frac{pr^2}{x^2} \]

### Step 4: Set the derivative to zero
Set the derivative equal to zero to find critical points:
\[ p - \frac{pr^2}{x^2} = 0 \]
This simplifies to:
\[ p = \frac{pr^2}{x^2} \]
Dividing both sides by \( p \) (assuming \( p \neq 0 \)):
\[ 1 = \frac{r^2}{x^2} \]
Thus, we have:
\[ x^2 = r^2 \implies x = r \quad (\text{since } x \text{ must be positive}) \]

### Step 5: Find the second derivative
To confirm that this is a minimum, we find the second derivative:
\[ f''(x) = \frac{2pr^2}{x^3} \]
Since \( p > 0 \) and \( r > 0 \), \( f''(x) > 0 \) for \( x = r \). This indicates that \( f(x) \) has a local minimum at \( x = r \).

### Step 6: Calculate the minimum value
Now, substitute \( x = r \) back into the expression to find the minimum value:
\[ f(r) = pr + \frac{pr^2}{r} = pr + pr = 2pr \]

Thus, the minimum value of \( px + py \) when \( xy = r^2 \) is:
\[ \boxed{2pr} \]