Browse Questions

# Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

$\begin{array}{1 1} 8\;and\;8 \\ 9 \;and\;7 \\ 10\;and\;6 \\ 11 \;and \;5 \end{array}$

Toolbox:
• $\large\frac{d}{dx}$$(x.y)=x.\large\frac{d}{dx}$$(y)+y.\large\frac{d}{dx}$$(x) Given Sum of the No. = 16 Let sum of cubes be S Let two numbers be x and 16-x Sum of cubes S=x^3+(16-x)^3 Differentiating with respect to x we get \large\frac{dS}{dx}$$=3x^2+3(16-x)^2.(-1)$
$\quad\;\;=3x^2-3(16-x)^2.(-1)$
$\quad\;\;=3(32x-256)$
Step 2:
For minimum value
$\large\frac{dS}{dx}$$=0 3(32x-256)=0 x=\large\frac{256}{32}$$=8$
Also $\large\frac{d^2S}{dx^2}$$=96>0$
$\Rightarrow S$ is minimum at $x=8$
Hence the required numbers are $8$ & $(16-8)\Rightarrow$ 8 & 8
edited Aug 20, 2013