1) Part the number 100 into two which the sum of their cubes is minimum.
Okay, so we have 1 variable here:
100 = a + (100 - a)
And we're trying to minimize:
a^3 + (100 - a)^3
In the domain [0, 100]
We get the derivative of this:
d/dx (a^3 + (100-a)^3) = 3a^2 - 3*(100-a)^2
Simplifies into:
600 (a - 50)
Find where this is zero, in this case a = 50. This is where the minimum is. I forget what theorem this comes from.
So your minimum is 250,000 for the sum of cubes.