1. ## Lagrange multiplier

Decide, using Lagrange multiplier, the extremes for
$\displaystyle f(x,y)=z^3+yz$
where
$\displaystyle y+3z+24=0$

I've tried, but I just couldn't solve the system of equations that arise. Maybe it's z^3 that confuses me.

2. ## Re: Lagrange multiplier

Do you mean $\displaystyle f(y,z)$ ?

4. ## Re: Lagrange multiplier

how do you do it?

5. ## Re: Lagrange multiplier

Attempt:
$\displaystyle 3\cdot \frac{\partial L}{\partial y}-\frac{\partial L}{\partial z}=0\; \Rightarrow\; 3z\left( 1-z \right)=y\; \Rightarrow\; \left( 1,0 \right)\; and\; \left( 0,0 \right).$

The book says $\displaystyle \left( 4,-36 \right)\; and \; \left( -2,-18 \right)$

6. ## Re: Lagrange multiplier

We are asked to used Lagrange multipliers to find the extrema for:

$\displaystyle f(y,z)=z^3+yz$

subject to the constraint:

$\displaystyle g(y,z)=y+3z+24=0$

By the theorem of Lagrange, we need to solve the system:

$\displaystyle z=\lambda$

$\displaystyle 3z^2+y=3\lambda$

These two equations imply:

$\displaystyle y=3z-3z^2$

Substituting for y into the constraint yields:

$\displaystyle z^2-2z-8=0$

$\displaystyle (z-4)(z+2)=0$

Hence, the extrema occur at:

$\displaystyle (y,z)=(-18,-2),\,(-36,4)$

thanks!