Find two positive numbers whose sum is 16 and the sum of whose cu… - Vidyadip

Question

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

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Answer

Let one number be x . Then, the other number is $(16-\mathrm{x})$.
Let $S(x)$ be the sum of these number. Then,
$S(x)=x^3+(16-x)^3$
$\Rightarrow S^{\prime}(x)=3 x^2-3(16-x)^2$
$\Rightarrow S^{\prime \prime}(x)=6 x+6(16-x)$
Now, $S^{\prime}(x)=0$
$\Rightarrow 3 x^2-3(16-x)^2=0$
$\Rightarrow x^2-(16-x)^2=0$
$\Rightarrow x^2-256-x^2+32 x=0$
$\Rightarrow x=8$
Now, $S^{\prime \prime}(8)=6(8)+6(16-8)$
$=48+48=96>0$
Then, by second derivative test, $x=8$ is the point of local minima of $S$.
Therefore, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16-8=8$.

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