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(D) Let the two factors of $36$ be $x$ and $\frac{36}{x}$.Let the sum of the factors be $S(x) = x + \frac{36}{x}$.To find the minimum sum,we different

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None of these

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(D) Let the two factors of $36$ be $x$ and $\frac{36}{x}$.Let the sum of the factors be $S(x) = x + \frac{36}{x}$.To find the minimum sum,we differentiate $S(x)$ with respect to $x$:$S'(x) = 1 - \frac{36}{x^2}$.Setting $S'(x) = 0$ for critical points:$1 - \frac{36}{x^2} = 0 \implies x^2 = 36 \implies x = 6$ (since factors are positive).Checking the second derivative:$S''(x) = \frac{72}{x^3}$.At $x = 6$,$S''(6) = \frac{72}{216} > 0$,which confirms a minimum.The factors are $x = 6$ and $\frac{36}{6} = 6$.Thus,the factors are $6, 6$,which is not listed in the options $A, B,$ or $C$.

$36$ is factorized into two factors such that the sum of the factors is minimum. What are the factors?

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