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

The Correct Option is B

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 $16$.