If `x^(3) + 3 (ky)^(3) = k^(3)` and `(dy)/(dx) + (x^(2))/(24 y^(2)) = 0`, then k is (where k `gt` 0)

Options

A. `(3)/(2)`
B. 2
C. `(2)/(3)`
D. 3

Explanation

To solve the problem, we need to find the value of \( k \) given the equations:

1. \( x^3 + 3(ky)^3 = k^3 \)
2. \( \frac{dy}{dx} + \frac{x^2}{24y^2} = 0 \)

### Step 1: Differentiate the first equation

We start with the first equation:

\[ x^3 + 3(ky)^3 = k^3 \]

Differentiating both sides with respect to \( x \):

\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(3(ky)^3) = \frac{d}{dx}(k^3) \]

Using the power rule, we differentiate:

\[ 3x^2 + 3 \cdot 3(ky)^2 \cdot k \frac{dy}{dx} = 0 \]

This simplifies to:

\[ 3x^2 + 9k^2y^2 \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)

Rearranging the equation gives us:

\[ 9k^2y^2 \frac{dy}{dx} = -3x^2 \]

Dividing both sides by \( 9k^2y^2 \):

\[ \frac{dy}{dx} = -\frac{3x^2}{9k^2y^2} = -\frac{x^2}{3k^2y^2} \]

### Step 3: Compare with the second equation

Now we have:

\[ \frac{dy}{dx} = -\frac{x^2}{3k^2y^2} \]

From the second equation given in the problem:

\[ \frac{dy}{dx} + \frac{x^2}{24y^2} = 0 \]

This implies:

\[ \frac{dy}{dx} = -\frac{x^2}{24y^2} \]

### Step 4: Set the two expressions for \(\frac{dy}{dx}\) equal

Now we can set the two expressions for \(\frac{dy}{dx}\) equal to each other:

\[ -\frac{x^2}{3k^2y^2} = -\frac{x^2}{24y^2} \]

### Step 5: Simplify and solve for \( k \)

Since \( x^2 \) and \( y^2 \) are common in both sides, we can cancel them out (assuming \( x \neq 0 \) and \( y \neq 0 \)):

\[ \frac{1}{3k^2} = \frac{1}{24} \]

Cross-multiplying gives:

\[ 24 = 3k^2 \]

Dividing both sides by 3:

\[ k^2 = 8 \]

Taking the positive square root (since \( k > 0 \)):

\[ k = \sqrt{8} = 2\sqrt{2} \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{2\sqrt{2}} \]