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

4 answers

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?
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
dang those typos, yeah? 64 works much better.
dont' know where the 65 came from

simply change it to 64 and complete the solution,

you should get 48 and 16