Question

find the absolute maximum of 144(x^3)(y^2)(1-x-y) in the first quadrant of the xy-plane

Answer 1

I don't understand what you are saying... do I just derive with respect to x, and then plug it into the derivative with respect to y?

Answer 2

the answer comes to be 1/3 at (1/2, 1/3), but I don't know how to get there..

Answer 3

we need to get the region...
since we are in the first quadrant we have\[144x^3y^2(1-x-y)\ge0\]um... thinking how to graph that :P

Answer 4

maybe we can just ignore the boundary at the moment and try to find some relative extrema for now...

Answer 5

How did you get the gradient to equal that? I've been trying to work it out and i have no idea how you got a 4x in there and how y gained an exra power..

Answer 6

yeah I think that grad is wrong too...

Answer 7

but I'm eating and am too lazy to work it out lol

Answer 8

LOL, yeah, i'm trying to work this out, and I'm puzzled..

Answer 9

\[\left\{-144 x^2 y^2 (4 x+3
y-3),-144 x^3 y (2 x+3
y-2)\right\}
\]
Sorry, I started with the wrong function.

Answer 10

Oh alright, but what i'm still now seeing is how you got 4x, or 3y.

Answer 11

not*

Answer 12

we need to get the region... since we are in the first quadrant we have
144x3y2(1−x−y)≥0
um... thinking how to graph that :P
Try x =2 and y=2

Answer 13

So the function goes to -Infinity when x and y go to +Infinity

Answer 14

Before factoring the gradient looks like
\[\left\{432 x^2 y^2
(-x-y+1)-144 x^3 y^2,288
x^3 y (-x-y+1)-144 x^3
y^2\right\}
\]
Factor each component correctly.

Answer 15

You should find that the absolute maximum of f is
\[ \frac 1 3\]

Answer 16

Here is a graph of your function near where it attains its absoulte Maxinun

Answer 17

Thank you so much eliassaab! I'm fanning you :)

Answer 18

You are welcome.