The sum of two nonzero numbers is 4. The mini | vedclass

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(A) Let the two nonzero numbers be $x$ and $y$. Given that $x + y = 4$,which implies $y = 4 - x$.We want to minimize the sum of their reciprocals,$S =

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$1$

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(A) Let the two nonzero numbers be $x$ and $y$. Given that $x + y = 4$,which implies $y = 4 - x$.We want to minimize the sum of their reciprocals,$S = \frac{1}{x} + \frac{1}{y}$.Substituting $y = 4 - x$ into the expression for $S$,we get $S(x) = \frac{1}{x} + \frac{1}{4 - x}$.To find the minimum,we differentiate $S$ with respect to $x$: $S'(x) = -\frac{1}{x^2} + \frac{1}{(4 - x)^2}$.Setting $S'(x) = 0$,we get $\frac{1}{x^2} = \frac{1}{(4 - x)^2}$,which implies $x^2 = (4 - x)^2$.Solving for $x$,$x^2 = 16 - 8x + x^2$,so $8x = 16$,which gives $x = 2$.If $x = 2$,then $y = 4 - 2 = 2$.The sum of the reciprocals is $S = \frac{1}{2} + \frac{1}{2} = 1$.Since $S''(x) = \frac{2}{x^3} + \frac{2}{(4 - x)^3}$,at $x = 2$,$S''(2) = \frac{2}{8} + \frac{2}{8} = \frac{1}{2} > 0$,confirming a local minimum.Thus,the minimum value is $1$.

The sum of two nonzero numbers is $4$. The minimum value of the sum of their reciprocals is

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