Applications Of Derivatives Question 307
Question: x and y be two variables such that $ x>0 $ and $ xy=1 $ . Then the minimum value of $ x+y $ is
[Kurukshetra CEE 1998; MP PET 2002]
Options:
A) 2
B) 3
C) 4
D) 0
Show Answer
Answer:
Correct Answer: A
Solution:
$ xy=1\Rightarrow y=\frac{1}{x} $ and let $ z=x+y $
$ z=x+\frac{1}{x}\Rightarrow \frac{dz}{dx}=1-\frac{1}{x^{2}} $
Now $ \frac{dz}{dx}=0\Rightarrow 1-\frac{1}{x^{2}}=0 $
$ x=-1,,+1 $ and $ \frac{d^{2}z}{dx^{2}}=\frac{2}{x^{3}} $
$ {{( \frac{d^{2}z}{dx^{2}} )}_{x=1}}=\frac{2}{1}=2=+ive $
$ \therefore $
Hence $ x=1 $ is point of minima and $ x=1 $ and $ y=1 $
$ \therefore $ Minimum value $ =x+y=2 $ .
BETA
