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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives1 Mark
Question
Fill in the blanks:
The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is ______.
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Answer

The least value of the function $\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{a}}(\text{a}>0,\text{b}>0,\text{x}>0)$ is $2\sqrt{\text{ab}}$
Solution:
$\text{f(x)}=\text{ax}+\frac{\text{b}}{\text{x}^2}$
$\text{f(x)}=0$
$\Rightarrow\ \text{a}=\frac{\text{b}}{\text{x}^2}$ or $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}\ \ (\text{as x}>0)$
Now, $\text{f}''(\text{x})=\frac{2\text{b}}{\text{x}^3}>0$ for $\text{x}=\sqrt{\frac{\text{b}}{\text{a}}}$
Thus least value of f(x) is $\text{f}\Big(\sqrt{\frac{\text{b}}{\text{a}}}\Big)=\text{a}\cdot\sqrt{\frac{\text{b}}{\text{a}}}+\frac{\text{b}}{\sqrt{\frac{\text{b}}{\text{a}}}}$ $=\sqrt{\text{ab}}+\sqrt{\text{ab}}=2\sqrt{\text{ab}}$

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