Question

Two numbers have a sum of 10. find the minimum value that the sum of their cubes could be

Answer 1

Let
`x` be one of the numbers. Then the other one is
`10-x`.

Therefore, the sum of their cubes is:

`x^3+(10-x)^3=x^3+1000-300x+30x^2-x^3=30x^2-300x+1000`

Now we need to find the minimum value of the resulting function.

`(30x^2-300x+1000)'=60x-300`

`60x-300=0\\60x=300\\x=5`

The derivative is negative to the left of
`5` and positive to the right of it, therefore, at
`5` there exists the minimum.

`5^3+(10-5)^3=125+125=250`

Therefore, the minimum value is 250.