divide 64 into two parts such that the product of one of them plus the cube of the other is a maximum

Answer 1

x+y=64

z = xy^3 = (64-y)y^3 = 64y^3 - y^4
dz/dy = 192y^2 - 4y^3 = 4y^2(48-y)

I think the way is now clear, right?

Answer 2

A product is the result of a multiplication.
"product of one of them plus the cube of the other" is mathematically confusing, since "plus" implies addition.
I will assume you meant:
"product of one of them with the cube of the other"

let the one to be cubed be x
then the other is 65-x

let S be the product of one of them plus the cube of the other
S = x^3(65-x) = 65x^3 - x^4

dS/dx = 195x^2 - 4x^3
= 0 for a max of S

195x^2 - 4x^3 = 0
x^2(195 - 4x) = 0
x^2 = 0 ---> x = 0
then the product as stated would be zero, not very big

or

4x = 195
x = 195/4

one is 195/4 , the other is 65 -195/4 or 65/4
as decimals:
48.75 and 16.25

Answer 3

dang those typos, yeah? 64 works much better.

Answer 4

dont' know where the 65 came from

simply change it to 64 and complete the solution,

you should get 48 and 16