What is the minimum value of `a^(2)x + b^(2)y`, if `xy = c^(2)`

Options

A. abc
B. 2abc
C. 3abc
D. 4abc

Answer

To find the minimum value of \( P = a^2 x + b^2 y \) given the constraint \( xy = c^2 \), we can use the method of substitution and calculus. Here’s the step-by-step solution:

### Step 1: Substitute for \( y \)
Given the constraint \( xy = c^2 \), we can express \( y \) in terms of \( x \):
\[ y = \frac{c^2}{x} \]

### Step 2: Rewrite \( P \)
Substituting \( y \) into the expression for \( P \):
\[ P = a^2 x + b^2 \left(\frac{c^2}{x}\right) = a^2 x + \frac{b^2 c^2}{x} \]

### Step 3: Differentiate \( P \)
Now, we differentiate \( P \) with respect to \( x \):
\[ \frac{dP}{dx} = a^2 - \frac{b^2 c^2}{x^2} \]

### Step 4: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ a^2 - \frac{b^2 c^2}{x^2} = 0 \]
This simplifies to:
\[ a^2 = \frac{b^2 c^2}{x^2} \]

### Step 5: Solve for \( x \)
Rearranging gives:
\[ x^2 = \frac{b^2 c^2}{a^2} \]
Taking the square root:
\[ x = \frac{bc}{a} \]

### Step 6: Find \( y \)
Now, substitute \( x \) back into the equation for \( y \):
\[ y = \frac{c^2}{x} = \frac{c^2}{\frac{bc}{a}} = \frac{ac}{b} \]

### Step 7: Substitute \( x \) and \( y \) back into \( P \)
Now we can find the minimum value of \( P \):
\[ P = a^2 \left(\frac{bc}{a}\right) + b^2 \left(\frac{ac}{b}\right) \]
This simplifies to:
\[ P = abc + abc = 2abc \]

### Conclusion
Thus, the minimum value of \( P = a^2 x + b^2 y \) under the constraint \( xy = c^2 \) is:
\[ \boxed{2abc} \]