Multivariable Maximum problem

**guess** · March 29th 2010, 09:00 AM

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.

**annie** · March 29th 2010, 09:10 AM

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????

**HallsofIvy** · March 29th 2010, 11:50 AM

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 $<y+ z- 4yz, x+ z- 4xz, x+ y- 4xy>= \lambda <1, 1, 1>$ . 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.

**guess** · March 29th 2010, 07:06 PM

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

**simplependulum** · March 29th 2010, 07:26 PM

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 .

**HallsofIvy** · March 30th 2010, 02:36 AM

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!

Thread: Multivariable Maximum problem

LinkBack

Thread Tools

Display

Multivariable Maximum problem

its urgent

Thanks

Similar Math Help Forum Discussions

maximum value of a multivariable function ?

Multivariable Differentiation Problem

Absolute maximum and minimum, Multivariable

Maximum and Minimum of multivariable calc

Maximum value of a multivariable function

Search Tags