1. ## Multivariable Maximum problem

Find all the points (x, y, z) with x, y, z => 0 which maximizes the function
f(x, y, z) = xy + xz + yz - 4xyz subject to the constraint x + y + z = 1.

2. ## its urgent

hi!!!
please help me n answer my questions???? what are the effects of rearranging and regrouping series? their difference and some examples. what about convergence after regrouping and rearranging????

3. Please do NOT "hijack" other peoples threads by asking completely unrelated questions in them! It is extremely rude.

guess, use the "Lagrange Multiplier Method"- the max or min of f(x,y,z) subject to the constraint g(x,y,z)= constant must satisfy $\nabla f= \lambda \nabla g$ for some number $\lambda$.

For this problem, that is $= \lambda <1, 1, 1>$. That is, you have $y+ z- 4yz= \lambda$, $x+ z- 4xz= \lambda$ and $x+ y- 4xy= \lambda$.

Those fairly easily reduce to y+ z- 4yz= x+ z- 4xz and y+ z- 4yz= x+ y- 4xy. Together with x+ y+ z= 1, you have three equations to solve for x, y, and z.

4. ## Thanks

Hi hallofivy,

I have tried your method too. However, it does not return the absolute extremum value.

when i re try it, the partial derviative of x and y is not equal to zero

5. well , i am not keen on inequalities but i think it is a great attempt !

$xy + yz + zx - 4xyz$

$= \frac{ (1-4x)(1-4y)(1-4z) - 1 + 4(x+y+z) }{16}$

$= \frac{ (1-4x)(1-4y)(1-4z) + 3 }{16}$

$= \frac{1}{16} \left[ 3 + (1-4x)(1-4y)(1-4z) \right]$

use $GM \leq AM$

$\leq \frac{1}{16} \left[ 3 + \left( \frac{ (1-4x ) + (1-4y ) + (1-4z ) }{3} \right )^3\right]$

$= \frac{1}{16} ( 3 - \frac{1}{27} ) = \frac{5}{27}$

the equality holds when $x =y =z = 1/3$

but it may be wrong because we can find that :

$GM \leq AM \leq 0$ !!

I hope it will be perfectly corrected by some experts .

6. Originally Posted by guess
Hi hallofivy,

I have tried your method too. However, it does not return the absolute extremum value.

when i re try it, the partial derviative of x and y is not equal to zero
The partial derivatives wrt x and y, for a constrained max or min do not have to be 0!