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

Solution :

Given function :

$$f(x)=\frac{5x^2}{2}+\frac{\alpha}{x^5}, \quad x>0$$

For minimum value :

$$f'(x)=0$$

Differentiate :

$$f'(x)=5x-\frac{5\alpha}{x^6}$$

Setting derivative equal to zero :

$$5x-\frac{5\alpha}{x^6}=0$$

$$5x=\frac{5\alpha}{x^6}$$

$$x^7=\alpha$$

$$x=\alpha^{1/7}$$

Second derivative :

$$f''(x)=5+\frac{30\alpha}{x^7}$$

Since for $$x>0$$ and $$\alpha>0$$,

$$f''(x)>0$$

therefore the function has minimum at :

$$x=\alpha^{1/7}$$

Now substitute in $$f(x)$$ :

$$f(\alpha^{1/7})=\frac{5(\alpha^{1/7})^2}{2}+\frac{\alpha}{(\alpha^{1/7})^5}$$

$$=\frac{5\alpha^{2/7}}{2}+\alpha^{2/7}$$

$$=\alpha^{2/7}\left(\frac{5}{2}+1\right)$$

$$=\frac{7\alpha^{2/7}}{2}$$

Given minimum value is :

$$14$$

Therefore,

$$\frac{7\alpha^{2/7}}{2}=14$$

$$\alpha^{2/7}=4$$

$$\alpha=4^{7/2}$$

$$=(2^2)^{7/2}$$

$$=2^7$$

$$=128$$

Final Answer :

$$128$$