Question

Find two positive numbers whose product is $64$ and the sum is minimum.

Answer 1

Let two positive numbers $xy=64$
According to question $xy=64$
$\therefore y=\frac{64}{x}$
sum of two positive numbers $z=x+y$
$s=x+\frac{64}{x}$
differentiate with respect to $x$
$\frac{ds}{dx}=\frac{d}{dx}(x+\frac{64}{x})=1-\frac{64}{x^2}....\left(i\right)$
$\frac{ds}{dx}=0$
$1-\frac{64}{x^2}=0\Rightarrow x^2=64\Rightarrow x=8$
differentiate the equation (1) again
$\frac{d^{2}s}{d{x}^2}=0+\frac{128}{x^2}$
$\therefore$ at point $x=8,\frac{d^{2}s}{d{x}^2}=\frac{128}{8^2}=\frac{1}{4}>0$
$\therefore$ value of $s$ minimum at point $x=8$
$y=\frac{64}{x}=\frac{64}{8}=8$
$\therefore$ required numbers are $8,8$