Applications Of Derivatives Question 352
Question: If $ ab=2a+3b,,a>0,b>0 $ then the minimum value of ab is
[Orissa JEE 2002]
Options:
A) 12
B) 24
C) $ \frac{1}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ ab=2a+3b $
therefore $ (a-3)b=2a $
therefore $ b=\frac{2a}{a-3} $ N
Now $ z=ab=\frac{2a^{2}}{a-3} $
therefore $ \frac{dz}{da}=\frac{2[(a-3)2a-a^{2}]}{{{(a-3)}^{2}}}=\frac{2[a^{2}-6a]}{{{(a-3)}^{2}}} $
Put $ \frac{dz}{da}=0 $ , \ $ a^{2}-6a=0 $ , $ a=0,,6 $ Now at $ a=6, $
$ \frac{d^{2}z}{da^{2}}=+ve $
When $ a=6,b=4 $ ; \ (ab)min. = 6 × 4 = 24.
BETA
