Question

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

Answer 1

Let one number be$x$. Then, the other number is $(16 − x)$.

Let the sum of the cubes of these numbers be denoted by $S(x)$. Then,

$s(x)=x^{3}+(16-x)^{3}$

$s'(x)=3x^{2}-3(16-x)^{2},s"(x)=6x+6(16-x)$

Now,$s''(x)=0=3x^{2}-3(16-x)^{2}=0$

$x^{2}-(16-x)^{2}=0$

$x^{2}-256-x^{2}+32x=0$

$x=\frac{256}{32}=8$

$s''(8)=6(8)+6(16-8)=48+48=96>0$

Now,

∴ By second derivative test, $x = 8$ is the point of local minima of S.

Hence, the sum of the cubes of the numbers is the minimum when the numbers are 8 and $16 − 8 = 8$