Question

find the two numbers whose sum is 12 and sum of whose cubes is minimum

Answer 1

let the two parts be x and y.
x+y = 12
⇒ y = (12-x)

You need to minimise the sum of their cubes.
(x)³ = x³
(12-x)³ = (12)³ - x³ + 3×12×x² - 3×x×12² = 1728 - x³ + 36x² - 432x

sum, s = (1728 - x³ + 36x² - 432x) + x³ = 1728 + 36x² - 432x
⇒ s = 36x² - 432x + 1728

$\frac{ds}{dx}=0\\ \\ \Rightarrow \frac{d}{dx}(36 x^{2} -432x+1728)=0 \\ \\ \Rightarrow 36 \times 2x - 432+0 =0\\ \\ \Rightarrow 72x-432=0\\ \\ \Rightarrow 72x=432\\ \\ \Rightarrow x= \frac{432}{72} \\\\\Rightarrow x=6$

For minimum, $\frac{d^2s}{dx^2} \ \ \textgreater \ \ 0$

$\frac{d^2s}{dx^2}= \frac{d^2(36x^2-432x+1728)}{dx^2}= \frac{d}{dx}(72x-432)=72\ \textgreater \ 0\\ \\ (condition\ satisfied)$

So x = 6,
y = 12 - x = 12- 6 = 6

So the two parts are (6,6)