Multivariable max&min

Mar 2010
165
40
Planet earth.
Find the max&min of the following function :
\(\displaystyle f(x,y,z)=8xy-3y^2+32z+5\)
subject to the constraint : \(\displaystyle 4x^2+y^2+4z^2=24\)

By the Lagrang Multipliers' method :

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

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

then ?!!
 
Jul 2007
894
298
New Orleans
Find the max&min of the following function :
\(\displaystyle f(x,y,z)=8xy-3y^2+32z+5\)
subject to the constraint : \(\displaystyle 4x^2+y^2+4z^2=24\)

By the Lagrang Multipliers' method :

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

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

then ?!!

sub \(\displaystyle y = \lambda x\)

into

\(\displaystyle 4x- 3y = \lambda y\)

and get

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

factor out and x

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

x= 0

Now plug that x = 0 into

\(\displaystyle y = \lambda x\)

and get that y = 0

Last plug x = y = 0

into

\(\displaystyle 4x^2 +y^2 +4z^2 =24\)
 
Mar 2010
165
40
Planet earth.
thanks !!

But why did you ignore solving for \(\displaystyle \lambda\) in :

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

??
 
Jan 2010
564
242
Kuwait
By using the method of Lagrange, we will get:

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

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

when x=0 ---> y=0 , substitute this in 4 ---> \(\displaystyle z^2=6 \implies z=\pm \sqrt{6}\)
so the first two points are \(\displaystyle (0,0,\sqrt{6})\) and \(\displaystyle (0,0,-\sqrt{6})\)
when \(\displaystyle \lambda = 1\) ---> z=4 from (3) ---> y=x from (1) and (2)
So when \(\displaystyle \lambda=1\) we have y=x and z=4 , By substituting this in (4) :
\(\displaystyle 5x^2+64=24 \implies x^2=-8\) which don't have any solutions
so when \(\displaystyle \lambda = 1\), we do not have any points.
when \(\displaystyle \lambda=-4\) ---> y=-4x from (1) and z=-1 from (3)
by substituting this in (4) ---> \(\displaystyle x^2=1 \implies x=\pm 1\)
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 : \(\displaystyle (0,0,\sqrt{6}),(0,0,-\sqrt{6}),(1,-4,-1)\) and \(\displaystyle (-1,4,-1)\)
Find the value of the function at these point, biggest --> max, smallest --> min.
 
Jul 2007
894
298
New Orleans
thanks !!

But why did you ignore solving for \(\displaystyle \lambda\) in :

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

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