# Thread: Lagrange Multiplier

1. ## Lagrange Multiplier

I'm stuck on a pretty straightforward LM type of problem.

Here's the problem statement.
a,b,c are positive constants. x,y,z are positive and that ayz+bzx+cxy=3abc. Show that xyz =< abc.

So I've approached it by setting f(x,y,z)=xyz subject to ayz+bzx+cxy=3abc

so F(x,y,z,lamda)=xyz-lamda(ayz+bzx+cxy-3abc)
then I kinda got lost...

Any pointers? Thanks.

2. Originally Posted by geegee
I'm stuck on a pretty straightforward LM type of problem.

Here's the problem statement.
a,b,c are positive constants. x,y,z are positive and that ayz+bzx+cxy=3abc. Show that xyz =< abc.

So I've approached it by setting f(x,y,z)=xyz subject to ayz+bzx+cxy=3abc

so F(x,y,z,lamda)=xyz-lamda(ayz+bzx+cxy-3abc)
then I kinda got lost...

Any pointers? Thanks.
I think what you did is good. Why don't you find the maximum of $\displaystyle F(x,y,z,\lambda)$? Take the partial derivatives of $\displaystyle F$ with respect to $\displaystyle x$, $\displaystyle y$ and $\displaystyle z$ and equal them to $\displaystyle 0$. Then solve for each variable, except maybe for $\displaystyle \lambda$ which doesn't really matter.

3. You should obtain (L is lambda)

1) yz/(bz+cy)= L

2) xz/(az+cx) = L

3) xy/(ay+bx) = L

Using 1 and 2 x = (a/b)y

Using 2 and 3 z = (c/b)y

Use these results in ayz+bzx+cxy=3abc

You should obtain y = +b

Then x = +a

and z = +c

The max of f = xyz then is abc