Lagrange Multipliers

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November 15th 2010, 04:07 PM
Playthious

Lagrange Multipliers

Find the maximum value of 2x + 2y + z on the sphere of radius 1 at the origin.
November 15th 2010, 04:11 PM
dwsmith

So your constraint is $\diaplystyle x^2+y^2+z^2=1$

Now find: $\displaystyle \triangledown f \ \mbox{and} \ \lambda\triangledown g$

Then set the i, j, and k components equal to each other of f and g where f is the function and g is the constraint.
November 15th 2010, 04:31 PM
Playthious

I've found lambda(X) = 1
lambda(Y) = 1
lambda(Z) = 1/2
and x^2 + y^2 + z^2 = 1
how would I solve this for a maximum, taking into account the case when lambda is 0?
November 15th 2010, 04:39 PM
dwsmith

I am just typing this up so I can see it.

$\displaystyle \triangledown f=2i+2j+k \ \mbox{and} \ \lambda\triangledown g=\lambda 2xi+\lambda 2yj+\lambda 2zk$

$\displaystyle 2=2x\lambda\rightarrow \lambda=\frac{1}{x}$

$\displaystyle 2=2y\lambda\rightarrow y=x$

$\displaystyle 1=2z\lambda\rightarrow z=\frac{x}{2}$

$\displaystyle x^2+x^2+\left(\frac{x}{2}\right)^2=1\rightarrow \frac{9x^2}{4}=1\rightarrow x=\pm \frac{2}{3}$

Since we want the max, we take the positive value.

$\displaystyle x=y=\frac{2}{3} \ \mbox{and} \ z=\frac{2}{6}$

Now plug into f.
November 15th 2010, 05:12 PM
Playthious

ok
thanks for your help
November 16th 2010, 07:00 AM
HallsofIvy

Since the value of $\lambda$ is not part of the solution, you can often simplify by first eliminating $\lambda$ by dividing one equation by another.
Here, you have $2= 2\lambda x$ and $2= 2\lambda y$ so, dividing the first by the second,
$\frac{2}{2}= \frac{2\lambda x}{2\lambda y}$ so that
$1= \frac{x}{y}$ or y= x. Similarly, dividing $2= 2\lambda x$ by $1= 2\lambda z$ gives x= 2z.

Putting those into $x^2+ y^2+ z^2= 1$ gives $4z^2+ 4z^2+ z^2= 9z^2= 1$ .