Let a number be x then other number is (16 - x).
Let the sum of their cubes is S.
Then ` S = x^(3) + (16 - x)^(3)`
` rArr (dS)/(dx) = 3x^(2)+3(16-x)^(2)(-1)`
` = 3x^(2) - 3(16-x)^(2)`
and ` ((d^(2)S)/(dx^(2))) = 6x + 6(16-x) = 96`
For minimum value ` (dS)/(dx) = 0`
` rArr 3x^(2) -3 (16-x)^(2)= 0`
` rArr x^(2) - (256+x^(2)-32x)=0`
` rArr 32x = 256 rArr x = 8`
at ` x = 8, ((d^(2)S)/(dx^(2)))_(x=8) = 96 gt 0 `
` :. ` at x = 8, S is minimum.
`:. 16 - x = 16 - 8 = 8 `
Therefore numbers are 8 and 8.
