# Multivariable max&min

#### Miss

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 ?!!

#### 11rdc11

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

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

#### Miss

thanks !!

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

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

??

#### General

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.

#### 11rdc11

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.