How should we choose two numbers, each greater than or equal to `-2` , whose sum is `1//2` so that the sum of the first and the cube of the second is minimum?

Explanation

Let `a` and `b` be two number such that
` a, b ≥ 2 `
Given as `a + b = 1/2`
Assume that `S = a+b^3`
`=>S=a+(1/2-a)^3`
`(dS)/(da)=1+3(1/2-a)^2(-1)`
condition maxima and minima is,
`(ds)/(da)=0`
`=>1+3(1/2-a)^2(-1)=0`
`=>3(1/2-a)^2=1`
`=>(1/2-a)^2=1/3`
`=>(1/2-a)=pm1/sqrt3`
`=>a=1/2pm1/sqrt3`
`(d^2S)/(da^2)=6(1/2-a)`
For S to minimum,`(d^2S)/(da^2) gt 0`
`=>a = 1/2-1/sqrt3`
hence,
`a=1/2-1/sqrt3` and `b=1/sqrt3`