Let `ab=1`, then the minimum value of `(1)/(a^(4))+(1)/(4b^(4))` is

Options

A. `1`
B. `2`
C. `1//4`
D. `1//2`

Explanation

To find the minimum value of the expression \(\frac{1}{a^4} + \frac{1}{4b^4}\) given that \(ab = 1\), we can use the concept of the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Substitute \(b\) in terms of \(a\)**:
Since \(ab = 1\), we can express \(b\) as:
\[ b = \frac{1}{a} \]

2. **Rewrite the expression**:
Substitute \(b\) into the expression:
\[ \frac{1}{a^4} + \frac{1}{4b^4} = \frac{1}{a^4} + \frac{1}{4\left(\frac{1}{a}\right)^4} = \frac{1}{a^4} + \frac{1}{4\cdot\frac{1}{a^4}} = \frac{1}{a^4} + \frac{a^4}{4} \]

3. **Let \(x = a^4\)**:
Now, we can rewrite the expression in terms of \(x\):
\[ f(x) = \frac{1}{x} + \frac{x}{4} \]

4. **Find the derivative**:
To find the minimum value, we differentiate \(f(x)\):
\[ f'(x) = -\frac{1}{x^2} + \frac{1}{4} \]

5. **Set the derivative to zero**:
Set \(f'(x) = 0\) to find critical points:
\[ -\frac{1}{x^2} + \frac{1}{4} = 0 \implies \frac{1}{4} = \frac{1}{x^2} \implies x^2 = 4 \implies x = 2 \quad (\text{since } x > 0) \]

6. **Evaluate the second derivative**:
To confirm that this is a minimum, we can check the second derivative:
\[ f''(x) = \frac{2}{x^3} \]
Since \(f''(x) > 0\) for \(x > 0\), this means \(f(x)\) is concave up at \(x = 2\), confirming a local minimum.

7. **Find the minimum value**:
Substitute \(x = 2\) back into \(f(x)\):
\[ f(2) = \frac{1}{2} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} = 1 \]

Thus, the minimum value of \(\frac{1}{a^4} + \frac{1}{4b^4}\) is \(\boxed{1}\).