AP EAMCET · Maths · Sequences and Series
If \(a^2 x^4+b^2 y^4=c^6\) then maximum value of \(x y=\)
- A \(\frac{c^3}{2 a b}\)
- B \(\frac{c^3}{\sqrt{2 a b}}\)
- C \(\frac{c^3}{a b}\)
- D \(\frac{c^3}{\sqrt{a b}}\)
Answer & Solution
Correct Answer
(B) \(\frac{c^3}{\sqrt{2 a b}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & a^2 x^4+b^2 y^4=c^6 \\ & \Rightarrow \frac{a^2 x^4+b^2 y^4}{2} \geq \sqrt{a^2 b^2 x^4 y^4} \Rightarrow \frac{c^6}{2} \geq a b x^2 y^2 \\ & \Rightarrow x^2 y^2 \leq \frac{c^6}{2 a b} \Rightarrow x y \leq \frac{c^3}{\sqrt{2 a b}} .\end{aligned}\)
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