If x and y are two positive real numbers such | vedclass

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(B) Given $xy = 4$, we can express $x$ as $x = \frac{4}{y}$.Let $f(y) = \sqrt{x} + \frac{y^2}{2} = \sqrt{\frac{4}{y}} + \frac{y^2}{2} = 2y^{-1/2} + \f

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$\frac{5}{2}$

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(B) Given $xy = 4$, we can express $x$ as $x = \frac{4}{y}$.Let $f(y) = \sqrt{x} + \frac{y^2}{2} = \sqrt{\frac{4}{y}} + \frac{y^2}{2} = 2y^{-1/2} + \frac{y^2}{2}$.To find the minimum, we differentiate $f(y)$ with respect to $y$:$f'(y) = 2(-\frac{1}{2})y^{-3/2} + \frac{1}{2}(2y) = -y^{-3/2} + y$.Setting $f'(y) = 0$ gives $y = y^{-3/2}$, which implies $y^{5/2} = 1$, so $y = 1$.For $y = 1$, $x = \frac{4}{1} = 4$.The value of the expression at $y = 1$ is $f(1) = \sqrt{4} + \frac{1^2}{2} = 2 + 0.5 = 2.5 = \frac{5}{2}$.Since $f''(y) = \frac{3}{2}y^{-5/2} + 1 > 0$ for $y > 0$, the function has a local minimum at $y = 1$.

If $x$ and $y$ are two positive real numbers such that $xy = 4$, then the minimum value of $\left(\sqrt{x} + \frac{y^2}{2}\right)$ is

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