If `a^2x^4+b^2y^4=c^6,` then the maximum value of `x y` is `(c^2)/(sqrt(a b))` (b) `(c^3)/(a b)` `(c^3)/(sqrt(2a b))` (d) `(c^3)/(2a b)`

Options

A. `(c^(2))/(sqrt(ab)`
B. `(c^(3))/(sqrt(ab)`
C. `(c^(3))/(sqrt(2ab)`
D. `(c^(3))/(sqrt(2ab)`

Explanation

To find the maximum value of \( xy \) given the equation \( a^2 x^4 + b^2 y^4 = c^6 \), we can use the method of Lagrange multipliers or substitution. Here, we'll use substitution to express \( y \) in terms of \( x \) and then maximize \( xy \).

### Step-by-Step Solution:

1. **Express \( y \) in terms of \( x \)**:
From the equation \( a^2 x^4 + b^2 y^4 = c^6 \), we can isolate \( y^4 \):
\[ b^2 y^4 = c^6 - a^2 x^4 \]
Dividing both sides by \( b^2 \):
\[ y^4 = \frac{c^6 - a^2 x^4}{b^2} \]
Taking the fourth root gives:
\[ y = \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{1/4} \]

2. **Formulate \( f(x) = xy \)**:
Substitute \( y \) into \( f(x) \):
\[ f(x) = x \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{1/4} \]

3. **Differentiate \( f(x) \)**:
To find the maximum, we differentiate \( f(x) \) with respect to \( x \) and set the derivative equal to zero:
\[ f'(x) = \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{1/4} + x \cdot \frac{1}{4} \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{-3/4} \cdot \left(-4a^2 x^3\right) \cdot \frac{1}{b^2} \]
Setting \( f'(x) = 0 \) gives us a critical point.

4. **Simplify the equation**:
After simplification, we find:
\[ \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{1/4} = \frac{a^2 x^3}{b^2} \cdot \left( \frac{c^6 - a^2 x^4}{b^2} \right)^{-3/4} \]

5. **Solve for \( x^4 \)**:
After some algebra, we find:
\[ 4x^4 = \frac{c^6}{2a^2} \]
Thus,
\[ x^4 = \frac{c^6}{8a^2} \]
Therefore,
\[ x = \left(\frac{c^6}{8a^2}\right)^{1/4} = \frac{c^{3/2}}{2^{1/2} a^{1/2}} \]

6. **Find \( y \)**:
Substitute \( x \) back to find \( y \):
\[ y^4 = \frac{c^6 - a^2 \left(\frac{c^6}{8a^2}\right)}{b^2} = \frac{c^6 \left(1 - \frac{1}{8}\right)}{b^2} = \frac{7c^6}{8b^2} \]
Thus,
\[ y = \left(\frac{7c^6}{8b^2}\right)^{1/4} = \frac{c^{3/2} \cdot 7^{1/4}}{2^{1/2} b^{1/2}} \]

7. **Calculate \( xy \)**:
Now, we can find \( xy \):
\[ xy = \left(\frac{c^{3/2}}{2^{1/2} a^{1/2}}\right) \left(\frac{c^{3/2} \cdot 7^{1/4}}{2^{1/2} b^{1/2}}\right) = \frac{c^3 \cdot 7^{1/4}}{2 a^{1/2} b^{1/2}} \]

8. **Determine the maximum value**:
The maximum value of \( xy \) is:
\[ \frac{c^3}{\sqrt{2ab}} \]

### Final Answer:
The maximum value of \( xy \) is \( \frac{c^3}{\sqrt{2ab}} \).