AP EAMCET · Maths · Application of Derivatives
The maximum value of \(x^4 y^4\) when \(a^2 x^4+b^2 y^4=c^6\) is
- A \(\frac{c^{12}}{16 a^4 b^4}\)
- B \(\frac{c^{12}}{4 a^2 b^2}\)
- C \(\frac{c^6}{(a+b)^{12}}\)
- D \(\frac{c^6}{a^4+b^4}\)
Answer & Solution
Correct Answer
(B) \(\frac{c^{12}}{4 a^2 b^2}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text {Let } f(x)=x^4 y^4 \\ & \text { And } a^2 x^4+b^2 y^4=c^6 \\ & \Rightarrow y^4=\frac{1}{b^2}\left(c^6-a^2 x^4\right) \\ & \therefore \quad f(x)=\mathrm{x}^4 \cdot \frac{1}{b^2}\left(c^6-a^2 x^4\right) \\ & =\frac{1}{b^2} x^4\left(c^6-a^2 x^4\right) \\ &…
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