Let x and y be two variables such that x 0 | vedclass

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(B) Let $A = x + y$. Since $xy = 1$,we have $y = \frac{1}{x}$.Substituting this into the expression for $A$,we get $A(x) = x + \frac{1}{x}$ for $x > 0

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$2$

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(B) Let $A = x + y$. Since $xy = 1$,we have $y = \frac{1}{x}$.Substituting this into the expression for $A$,we get $A(x) = x + \frac{1}{x}$ for $x > 0$.To find the minimum value,we find the derivative: $\frac{dA}{dx} = 1 - \frac{1}{x^2}$.Setting $\frac{dA}{dx} = 0$,we get $1 = \frac{1}{x^2}$,which implies $x^2 = 1$. Since $x > 0$,we have $x = 1$.Now,we check the second derivative: $\frac{d^2A}{dx^2} = \frac{2}{x^3}$.At $x = 1$,$\frac{d^2A}{dx^2} = \frac{2}{1^3} = 2 > 0$.Since the second derivative is positive,$x = 1$ is a point of local minima.The minimum value is $A(1) = 1 + \frac{1}{1} = 2$.

Let $x$ and $y$ be two variables such that $x > 0$ and $xy = 1$. What is the minimum value of $x + y$?

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