TS EAMCET · Maths · Application of Derivatives
If \(x\) and \(y\) are two positive real numbers such that \(x y=4\) then the minimum value of \(\left(\sqrt{x}+\frac{y^2}{2}\right)\) is
- A \(4\)
- B \(\frac{5}{2}\)
- C \(2 \sqrt{2}\)
- D \(\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{2}\)
Step-by-step Solution
Detailed explanation
Let \(f(x) = \sqrt{x}+\frac{y^2}{2}\). Given \(xy=4\), then \(y=\frac{4}{x}\). \(f(x) = \sqrt{x}+\frac{(4/x)^2}{2} = x^{1/2}+8x^{-2}\) \(f'(x) = \frac{1}{2}x^{-1/2}-16x^{-3}\) Set \(f'(x)=0\): \(\frac{1}{2\sqrt{x}} = \frac{16}{x^3} \Rightarrow x^{5/2} = 32 \Rightarrow x=4\) For…
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