Question

split 20 into 2 nonnegative numbers x,y such that the product of x and y^2 is a maximum

Answer 1

\[
x+y = 20 \\
\max(xy^2)
\]

Answer 2

\[
x,y\ge0
\]

Answer 3

Yes, I found out that the derivative of the primary is (sqrt 5)/(sqrt x) but when solving for x it is not possible?

Answer 4

Hmm...
Look at it this way
\[\large y = 20 -x\]
And find the maximum of
\[\large xy^2 = x(20-x)^2=x(x^2-40x+400)\]

Answer 5

Don't you have to solve for y for the constraint though?

Answer 6

we did, didn't we? And we got
y = 20-x

Answer 7

I believe the constraint is xy^2=20

Answer 8

No, that is what we are supposed to maximise. The constraint is x + y = 20

Answer 9

Wow I feel stupid, thank you so much!!