Question

Split 20 into two nonnegative numbers x; y such that the product of x and y2 is a maximum.

Answer 1

That's supposed to be y^2...
sorry!

Answer 2

It's an optimization problem, I think that the primary function is xy^2, but I get stuck after that...

Answer 3

...help, anyone? I'm kinda lost...

Answer 4

well, diophontus would just create 2 setups that sum to 20 to work with

Answer 5

(10+n) + (10-n) = 20

(10+n) (10-n)^2

Answer 6

if we take the derivative of that and set it to zero .... we get a value for n

Answer 7

um... okay? I think I see where you're going with this, but I'm supposed to be using calculus to solve these... like with derivatives and stuff? We just learned it today and I'm totally lost...

Answer 8

might have to swap the x and y parts tho .... but its a start

Answer 9

All help is appreciated! I tried doing it this way and my instructor said it was wrong....

Answer 10

im not sure what "this way" is?

Answer 11

i get a cubic, that derives to a quadratic, that is easily determined by a quadratic formula from algebra to determine "n"

Answer 12

...neither am I, believe me. She was talking about constraint and primary functions (hence the xy^2 I had earlier). I can't find the constraint function.

Answer 13

the constraint is x+y = 20

Answer 14

OH! Seriously?!! Ack, I had that the first time... thanks, I can handle it now!! (This is what lack of sleep does to a person >_<)

Answer 15

sounds like a ... Legrandge multipler ... forget how to spell it

Answer 16

im wondering if my diophontine was good :) i get 20/3 and 40/3

Answer 17

You are most certainly right!! I just had NO idea what you meant by "diophontine".

Answer 18

diophontus was an old greek guy that used to live down by the agora when i was a kid :)

Answer 19

if we do a Legendre multiplier we get:

\[F(x,y)=xy^2~:~L[G(x,y)=x+y-20]\]
\[Fx = y^2~:~L[Gx] = L\\Fy = 2xy~:~L[Gy] = L\]
equating F and G parts we get\[y^2=L\\y = \sqrt{L}\]

\[2xy = L\\x = \frac{1}{2y}L\\x = \frac{1}{2\sqrt{L}}L\\x = \frac12 \sqrt{L}\]


plugging x and y into G = 0 solves L,for the Legendre multiplier (still not too sure if i got the name right)

\[x+y=20\\\frac12\sqrt{L}+\sqrt{L}=20\\\frac32\sqrt{L}=20\\y = \sqrt{L}=\frac{40}{3}\\therefore ~x = \frac{20}{3}\]