The minimum value of the function $P(x)=K_1x+\frac{K_2}{x}, (x>0, K_1 > 0, K_2 > 0)$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $2\sqrt{K_1K_2}$
$P(x)=K_1x+\frac{K_2}{x}$
To find minimum value, $f'(x)=0$ and $f''(x)>0$.
$⇒P(x)=0$
$⇒K_1x+\frac{K_2}{x}=0$
$⇒x=±\sqrt{\frac{K_2}{K_1}}$
Now,
$P'(x)=\frac{2K_2}{x^3}=\frac{2K_2}{\sqrt{K_2}}×\sqrt{K_1}>0$
$∴P\left(\sqrt{\frac{K_2}{K_1}}\right)=K_1\sqrt{\frac{K_2}{K_1}}+K_2\sqrt{\frac{K_1}{K_2}}$
$=\sqrt{K_1}\sqrt{K_2}+\sqrt{K_2}+\sqrt{K_1}$
$=2\sqrt{K_1K_2}$