Question

Amongst all pairs of positive numbers with product $256$, find those whose sum is the least.

• A. $16$, $14$
• B. $16$, $16$
• C. $64$, $4$
• D. $32$, $8$

Answer 1

Answer: (B)

Let the required numbers be $x$ and $y$. Then,
$xy=256$ (given) $...(i)$
Let $S=x+y$. Then,
$S=x+\frac{256}{x}$ [Using $(i)$]
$\Rightarrow \frac{dS}{dx}=1-\frac{256}{x^2}$
and $\frac{d^{2}S}{d{x}^2}=\frac{512}{x^3}$
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
$\because \frac{d^{2}S}{d{x}^2}{|}_{x=-16}<0$
Now, ${\left(\frac{d^{2}S}{d{x}^2}\right)}_{x=16}=\frac{512}{{\left(16\right)}^3}$
$=\frac{1}{8}>0$
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 $16$.