The sum of two non-zero numbers is 4. The min | vedclass

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(C) Let the two non-zero numbers be $x$ and $y$. Given $x + y = 4$,so $y = 4 - x$.We want to minimize the sum of their reciprocals,$S = \frac{1}{x} +

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$1$

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(C) Let the two non-zero numbers be $x$ and $y$. Given $x + y = 4$,so $y = 4 - x$.We want to minimize the sum of their reciprocals,$S = \frac{1}{x} + \frac{1}{y}$.Substituting $y = 4 - x$,we get $S(x) = \frac{1}{x} + \frac{1}{4 - x} = \frac{4 - x + x}{x(4 - x)} = \frac{4}{4x - x^2}$.To find the minimum,we differentiate $S(x)$ with respect to $x$:$S'(x) = \frac{d}{dx} [4(4x - x^2)^{-1}] = -4(4x - x^2)^{-2} \cdot (4 - 2x) = \frac{-4(4 - 2x)}{(4x - x^2)^2} = \frac{8(x - 2)}{(4x - x^2)^2}$.Setting $S'(x) = 0$,we get $x - 2 = 0$,so $x = 2$.Since $x = 2$,then $y = 4 - 2 = 2$.The minimum value is $S(2) = \frac{1}{2} + \frac{1}{2} = 1$.

The sum of two non-zero numbers is $4$. The minimum value of the sum of their reciprocals is

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