A) \[c\sqrt{ab}\]
B) \[2c\sqrt{ab}\]
C) \[-c\sqrt{ab}\]
D) \[-2c\sqrt{ab}\]
Correct Answer: B
Solution :
\[xy={{c}^{2}}\] Þ \[y=\frac{{{c}^{2}}}{x}\] Þ \[f(x)=ax+by=ax+\frac{b{{c}^{2}}}{x}\] Differentiate with respect to x \[{f}'(x)=a-\frac{b{{c}^{2}}}{{{x}^{2}}}\] Put \[{f}'(x)=0\] Þ \[a{{x}^{2}}-b{{c}^{2}}=0\] Þ \[{{x}^{2}}=\frac{b{{c}^{2}}}{a}\] Þ \[x=\pm \,\,c\sqrt{b/a}\] At \[x=+\,c\sqrt{b/a,}\,\,ax+by\] will be minimum. The minimum value \[f\,\,\left( c\sqrt{\frac{a}{b}} \right)=a.c\sqrt{\frac{a}{b}}+\frac{b{{c}^{2}}}{c}.\sqrt{\frac{b}{a}}\] = \[2c\sqrt{ab}\].
You need to login to perform this action.
You will be redirected in
3 sec
