# Maxima proof

• Oct 28th 2009, 01:39 PM
Chris9620
Maxima proof
Hello!
I have a question I'm struggling with, my last question was answered so helpfully I thought I'd try my others out here! Especially since my exam is in less than 3 hours!

Its a 'Two variable function Extrema' question

Q: The gravitational attraction at the point (x,y)=(1/x)+(4/y)+(9/(4-x-y))
Prove that G(x,y) has a maximum value of 9

These type of questions always stump me, sorry.

Thanks for any aid
• Oct 31st 2009, 05:56 AM
Media_Man
Stump?
I can see why it stumped you. First, how did the exam go? Second, this function has no maximum (or minimum), local or absolute. (Surprised)

$G(x,y)=\frac1x+\frac4y+\frac9{4-x-y}$

Consider: $\lim_{x+y\to 4}G(x,y)=\pm\infty$ and $\lim_{x,y\to\infty}G(x,y)=0$ because of the $\frac9{4-x-y}$ term.

For traditional proof, $\bigtriangledown G=\vec{0}$ has no solution.