Multivariable max&min

**Miss** · May 16th 2010, 11:01 AM

Find the max&min of the following function :
$f(x,y,z)=8xy-3y^2+32z+5$
subject to the constraint : $4x^2+y^2+4z^2=24$

By the Lagrang Multipliers' method :

$8y=\lambda 8x$ ........(1)
$8x-6y=\lambda 2y$ .......(2)
$32=\lambda 8z$ .......(3)
$4x^2+y^2+4z^2=24$

(1) will be : $y=\lambda x$
(2) will be : $4x-3y=\lambda y$
(3) will be : $4=\lambda z$

then ?!!

**11rdc11** · May 16th 2010, 11:40 AM

Originally Posted by Miss

Find the max&min of the following function :
$f(x,y,z)=8xy-3y^2+32z+5$
subject to the constraint : $4x^2+y^2+4z^2=24$

By the Lagrang Multipliers' method :

$8y=\lambda 8x$ ........(1)
$8x-6y=\lambda 2y$ .......(2)
$32=\lambda 8z$ .......(3)
$4x^2+y^2+4z^2=24$

(1) will be : $y=\lambda x$
(2) will be : $4x-3y=\lambda y$
(3) will be : $4=\lambda z$

then ?!!

sub $y = \lambda x$

into

$4x- 3y = \lambda y$

and get

$-\lambda^2x+4x-3\lambda x = 0$

factor out and x

$x(-\lambda^2+4-3\lambda) = 0$

x= 0

Now plug that x = 0 into

$y = \lambda x$

and get that y = 0

Last plug x = y = 0

into

$4x^2 +y^2 +4z^2 =24$

**Miss** · May 16th 2010, 11:47 AM

thanks !!

But why did you ignore solving for $\lambda$ in :

$x(-\lambda^2+4-3\lambda) = 0$

??

**General** · May 16th 2010, 12:29 PM

By using the method of Lagrange, we will get:

$y=\lambda x$ ... (1)
$4x-3y=\lambda y$ ... (2)
$4=\lambda z$ ... (3)
$4x^2+y^2+4z^2=24$ ... (4)

By substituting (1) in (2), we will get the equation : $x(-\lambda^2 - 3\lambda + 4)=0$
$\implies x=0,\lambda=-4,\lambda=1$ ..

when x=0 ---> y=0 , substitute this in 4 ---> $z^2=6 \implies z=\pm \sqrt{6}$
so the first two points are $(0,0,\sqrt{6})$ and $(0,0,-\sqrt{6})$
when $\lambda = 1$ ---> z=4 from (3) ---> y=x from (1) and (2)
So when $\lambda=1$ we have y=x and z=4 , By substituting this in (4) :
$5x^2+64=24 \implies x^2=-8$ which don't have any solutions
so when $\lambda = 1$ , we do not have any points.
when $\lambda=-4$ ---> y=-4x from (1) and z=-1 from (3)
by substituting this in (4) ---> $x^2=1 \implies x=\pm 1$
when x=1 --> y=-4 ---> third point is (1,-4,-1)
when x=-1 ---> y=4 ---> fourth point is (-1,4,-1)
The points are : $(0,0,\sqrt{6}),(0,0,-\sqrt{6}),(1,-4,-1)$ and $(-1,4,-1)$
Find the value of the function at these point, biggest --> max, smallest --> min.

**11rdc11** · May 16th 2010, 01:12 PM

Originally Posted by Miss

thanks !!

But why did you ignore solving for $\lambda$ in :

$x(-\lambda^2+4-3\lambda) = 0$

??

Sorry forgot to copy that part from my worksheet when transfering it on here lol.

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