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Q.
Amongst all pairs of positive numbers with product $256$, find those whose sum is the least.
Application of Derivatives
Solution:
Let the required numbers be $x$ and $y$. Then,
$xy = 256$ (given) $\quad...(i)$
Let $S = x + y$. Then,
$S = x+\frac{256}{x}$ [Using $(i)$]
$\Rightarrow \frac{dS}{dx} = 1 - \frac{256}{x^{2}}$
and $ \frac{d^{2}S}{dx^{2}} = \frac{512}{x^{3}}$
For maximum or minimum values of $S$, we must have
$\frac{dS}{dx} = 0 \Rightarrow 1 - \frac{256}{x^{2}} = 0$
$\Rightarrow x^{2} =256$
$\Rightarrow x = 16$
$x = -16$ is neglected
$\because \frac{d^{2}S}{dx^{2}}\bigg|_{x = -16} < 0$
Now, $\left(\frac{d^{2}S}{dx^{2}}\right)_{x = 16} = \frac{512}{\left(16\right)^{3} }$
$ = \frac{1}{8} > 0$
Thus, $S$ is minimum when $x = 16$.
Putting $x = 16$ in $\left(i\right)$ we get $y = 16$.
Hence, the required numbers are both equal to $16$.