Applications Of Derivatives Question 336
Question: If $ xy=c^{2}, $ then minimum value of $ ax+by $ is
[RPET 2001]
Options:
A) $ c\sqrt{ab} $
B) $ 2c\sqrt{ab} $
C) $ -c\sqrt{ab} $
D) $ -2c\sqrt{ab} $
Show Answer
Answer:
Correct Answer: B
Solution:
$ xy=c^{2} $
therefore $ y=\frac{c^{2}}{x} $
therefore $ f(x)=ax+by=ax+\frac{bc^{2}}{x} $
Differentiate with respect to x $ {f}’(x)=a-\frac{bc^{2}}{x^{2}} $ Put $ {f}’(x)=0 $
therefore $ ax^{2}-bc^{2}=0 $
therefore $ x^{2}=\frac{bc^{2}}{a} $
therefore $ x=\pm c\sqrt{b/a} $
At $ x=+,c\sqrt{b/a,}ax+by $ will be minimum.
The minimum value $ f( c\sqrt{\frac{a}{b}} )=a.c\sqrt{\frac{a}{b}}+\frac{bc^{2}}{c}.\sqrt{\frac{b}{a}} $ = $ 2c\sqrt{ab} $ .
BETA
