If xy =10, then minimum value of `12x^2 + 13y^2` is equal to :

Options

A. 15
B. `40sqrt39`
C. `3sqrt13`
D. `30sqrt13`

Answer

To find the minimum value of the function \( 12x^2 + 13y^2 \) given the constraint \( xy = 10 \), we can follow these steps:

### Step 1: Express \( y \) in terms of \( x \)
Given the constraint \( xy = 10 \), we can express \( y \) as:
\[ y = \frac{10}{x} \]

### Step 2: Substitute \( y \) in the function
Now, substitute \( y \) in the function \( f(x) = 12x^2 + 13y^2 \):
\[ f(x) = 12x^2 + 13\left(\frac{10}{x}\right)^2 \]
This simplifies to:
\[ f(x) = 12x^2 + 13\left(\frac{100}{x^2}\right) = 12x^2 + \frac{1300}{x^2} \]

### Step 3: Differentiate the function
To find the minimum value, we need to differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}\left(12x^2 + \frac{1300}{x^2}\right) \]
Using the power rule and the quotient rule, we get:
\[ f'(x) = 24x - \frac{2600}{x^3} \]

### Step 4: Set the derivative to zero
To find the critical points, set \( f'(x) = 0 \):
\[ 24x - \frac{2600}{x^3} = 0 \]
Rearranging gives:
\[ 24x = \frac{2600}{x^3} \]
Multiplying both sides by \( x^3 \) results in:
\[ 24x^4 = 2600 \]

### Step 5: Solve for \( x \)
Now, solve for \( x \):
\[ x^4 = \frac{2600}{24} = \frac{1300}{12} = \frac{325}{3} \]
Taking the fourth root gives:
\[ x = \left(\frac{325}{3}\right)^{1/4} \]

### Step 6: Find \( y \)
Using the value of \( x \), find \( y \):
\[ y = \frac{10}{x} = \frac{10}{\left(\frac{325}{3}\right)^{1/4}} = 10 \cdot \left(\frac{3}{325}\right)^{1/4} \]

### Step 7: Substitute \( x \) and \( y \) back into the function
Now substitute \( x \) and \( y \) back into the function to find the minimum value:
\[ f(x, y) = 12x^2 + 13y^2 \]
Calculating \( f(x) \):
\[ f\left(\left(\frac{325}{3}\right)^{1/4}, \frac{10}{\left(\frac{325}{3}\right)^{1/4}}\right) = 12\left(\frac{325}{3}\right)^{1/2} + 13\left(\frac{10}{\left(\frac{325}{3}\right)^{1/4}}\right)^2 \]

### Step 8: Simplify to find the minimum value
After substituting and simplifying, you will arrive at the minimum value of \( 12x^2 + 13y^2 \).

### Final Answer
The minimum value of \( 12x^2 + 13y^2 \) given \( xy = 10 \) is \( 40\sqrt{39} \).

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