If the minimum value of $$f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5}, x \gt 0$$, is 14, then the value of $$\alpha$$ is equal to

We need to find the value of $$\alpha such that the minimum value of f(x) = \frac{5x^2}{2} + \frac{\alpha}{x^5} for x \gt 0$$ is 14.

Since the location of the minimum occurs where the derivative vanishes, we differentiate $$f(x) with respect to x: f'(x) = 5x - \frac{5\alpha}{x^6}. Setting f'(x)=0 gives 5x = \frac{5\alpha}{x^6}, which simplifies to x^7 = \alpha and hence x = \alpha^{1/7}$$.

To confirm that this critical point corresponds to a minimum, we compute the second derivative: $$f''(x) = 5 + \frac{30\alpha}{x^7}. Since for x \gt 0 and \alpha \gt 0 we have f''(x) \gt 0, the function attains a minimum at x = \alpha^{1/7}$$.

Substituting $$x = \alpha^{1/7} back into f(x) yields f(\alpha^{1/7}) = \frac{5(\alpha^{1/7})^2}{2} + \frac{\alpha}{(\alpha^{1/7})^5} = \frac{5\alpha^{2/7}}{2} + \frac{\alpha}{\alpha^{5/7}} = \frac{5\alpha^{2/7}}{2} + \alpha^{2/7} = \alpha^{2/7}\left(\frac{5}{2} + 1\right) = \frac{7\alpha^{2/7}}{2}\,.$$

Setting this minimum value equal to 14 gives $$\frac{7\alpha^{2/7}}{2} = 14, so \alpha^{2/7} = 4. Therefore, \alpha = 4^{7/2} = (2^2)^{7/2} = 2^7 = 128.

Thus, the correct answer is Option C: 128.