Finding Extrema of f(x, y, z) = 2x^2 + 8y^2 + z^2 subject to

[missace31]

Q: Find the point (x*,y*,z*) that is at the minimum of the function f(x,y,z,)=2x[sup:1zgtpijz]2[/sup:1zgtpijz]+8y[sup:1zgtpijz]2[/sup:1zgtpijz]+z[sup:1zgtpijz]2[/sup:1zgtpijz] subject to the constraint equation g(x,y,z) = 6x+4y+4z = 0

A: I know the steps in finding maximum and minimum for a function with constraints (set the first derivitive equal to zero to get the critical points and take the second derivitive to find whether the point is a max or min). I have already computed the partial derivitives of the function and the constraints:

?g/?x=6 ?g/?y=4 ?g/?z=4

?f/?x=4x ?f/?y=16y ?f/?z=2z

Plugging these values into the equation:

dg=(?g/?x)dx+(?g/?y)dy+(?g/?z)dz=0 we get: 0=6dx+4dy+4dz.

Plugging the partial derivitives into the equation:

df=(?f/?x)dx+(?f/?y)dy+(?f/?z)dz=0

we get:

0=4xdx+16ydy+2zdz

I do not know, however, how to simplify that last equation given the other equations because we have 3 variables. (This is not for a calc class, but I am just starting Multivariable Calc this semester so I am trying to learn these methods as quick as possible.. however I get stuck like this sometimes).

 

[soroban]

Hello, missace31!

\(\displaystyle \text{Find the point }(x^*,y^*,z^*)\text{ at the minimum of the function: }\:f(x,y,z)\:=\:2x^2 +8y^2+z^2\)

\(\displaystyle \text{subject to the constraint equation: }\:g(x,y,z) \:= \:6x+4y+4z \:= \:0\)

You should read up on Lagrange Multipliers.
Here's a sketchy game plan . . .

\(\displaystyle \text{Set up a new function: }\;F(x,y,z,\lambda) \;=\;2x^2 + 8y^2 + z^2 + \lambda(6x + 4y + 4z)\)


\(\displaystyle \text{Set the }f\!our\text{ partial derivatives equal to zero and solve.}\)

. . \(\displaystyle \begin{array}{ccccc}\frac{\partial F}{\partial x} & = & 4x + 6\lambda x & = & 0 \\ \\ \frac{\partial F}{\partial y} & = & 16y + 4\lambda y & = & 0 \\ \\ \frac{\partial F}{\partial z} & = & 2z + 4\lambda z & = & 0 \\ \\ \frac{\partial F}{\partial\lambda} & = & 6x + 4y + 4z &=& 0 \end{array}\)


\(\displaystyle \text{Then there is a "second partials test" to determine the nature of the extrema.}\)

 

[missace31]

I attempted to execute the problem with Lagrange multipliers as it is taught in the book I am working with (which is a little different than the way you explained it). When I did it that way (setting the partial derivitive of the function equal to lambda times the partial derivitive of the constraint) I got x*=y*=z*=0. This seemed wrong to me, so I tried it the way you explained it, and I guess I'm still a little confused. If I were to solve the equations for the partial derivitie of the new function with the term "L" then I would get the following three equations: lambda=-2/3, lambda=-4, and lambda=-1/2, not only does this not satisfy the fourth partial derivitive equation, but it also does satisfy the fact that all values of lambda should be equal because it is a constant. What am I doing wrong here?!?

 

[soroban]

Hello, missace31!


We had:

. . \(\displaystyle \begin{array}{ccccccccccc}\frac{\partial F}{\partial x} & = & 4x + 6\lambda x & = & 0 & & \Rightarrow & 2x(2+3\lambda) &=& 0 & [1] \\ \\ \frac{\partial F}{\partial y} & = & 16y + 4\lambda y & = & 0 & & \Rightarrow & 4y(4 + \lambda) &=&0 & [2] \\ \\ \frac{\partial F}{\partial z} & = & 2z + 4\lambda z & = & 0 & & \Rightarrow & 2z(1 + 2\lambda) &=& 0 & [3] \\ \\ \frac{\partial F}{\partial\lambda} & = & 6x + 4y + 4z &=& 0 & & \Rightarrow & 3x + 2y + 2z &=&0 & [4] \end{array}\)


\(\displaystyle \text{You are correct . . . }(0,0,0)\text{ is the minimum point.}\)


\(\displaystyle \text{Think about it . . .}\)

. . \(\displaystyle \text{The function }f(x,y,z) \:=\:2x^2+8y^2 + z^2 \text{ is always }\geq 0\)

. . \(\displaystyle \text{The point must be on the plane }6x + 2y + 4z \:=\:0\text{, which passes through the origin.}\)

\(\displaystyle \text{Therefore, the }origin\text{ produces the minimum value of }f(x,y,z).\)
. . \(\displaystyle \text{And the value of }\lambda\text{ is irrelevant.}\)

 

[Deleted member 4993]

I attempted to execute the problem with Lagrange multipliers as it is taught in the book I am working with (which is a little different than the way you explained it). When I did it that way (setting the partial derivitive of the function equal to lambda times the partial derivitive of the constraint) I got x*=y*=z*=0. This seemed wrong to me, so I tried it the way you explained it, and I guess I'm still a little confused. If I were to solve the equations for the partial derivitie of the new function with the term "L" then I would get the following three equations:

lambda=-2/3,

lambda=-4, and

lambda=-1/2

How did you get these values of 'L' ?

remember - from the equation

4x + 6Lx = 0 - you cannot just factor 'x' out - because then you are assuming 'x' is not equal to zero.