Applications Of Derivatives Question 337
Question: If $ a^{2}x^{4}+b^{2}y^{4}=c^{6}, $ then maximum value of xy is
[RPET 2001]
Options:
A) $ \frac{c^{2}}{\sqrt{ab}} $
B) $ \frac{c^{3}}{ab} $
C) $ \frac{c^{3}}{\sqrt{2ab}} $
D) $ \frac{c^{3}}{2ab} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ a^{2}x^{4}+b^{2}y^{4}=c^{6} $
therefore $ y={{( \frac{c^{6}-a^{2}x^{4}}{b^{2}} )}^{1/4}} $
Hence $ f(x)=xy=x{{( \frac{c^{6}-a^{2}x^{4}}{b^{2}} )}^{1/4}} $
therefore $ f(x)={{( \frac{c^{6}x^{4}-a^{2}x^{8}}{b^{2}} )}^{1/4}} $
Differentiate $ f(x) $ with respect to x, then $ {f}’(x)=\frac{1}{4}{{( \frac{c^{6}x^{4}-a^{2}x^{8}}{b^{2}} )}^{-3/4}}( \frac{4x^{3}c^{6}}{b^{2}}-\frac{8x^{7}a^{2}}{b^{2}} ) $ Put $ {f}’(x)=0 $ , $ \frac{4x^{3}c^{6}}{b^{2}}-\frac{8x^{7}a^{2}}{b^{2}}=0 $
therefore $ x^{4}=\frac{c^{6}}{2a^{2}} $
therefore $ x=\pm \frac{{c^{3/2}}}{{2^{1/4}}\sqrt{a}} $
At $ x=\frac{{c^{3/2}}}{{2^{1/4}}\sqrt{a}} $ the $ f(x) $ will be maximum, so
$ f( \frac{{c^{3/2}}}{{2^{1/4}}\sqrt{a}} ),=,{{( \frac{c^{12}}{2a^{2}b^{2}}-\frac{c^{12}}{4a^{2}b^{2}} )}^{1/4}}={{( \frac{c^{12}}{4a^{2}b^{2}} )}^{1/4}}=\frac{c^{3}}{\sqrt{2ab}} $ .
BETA
