# Thread: Help finding the extreme values

1. ## Help finding the extreme values

Hello,

I need to use the lagrange method to find the extreme values of
$f(y,z)=z^3+yz$

with this condition

$y+3z+24=0$

I got this system of equations
$z-\lambda=0$ (1)
$3z^2+y-3\lambda=0$ (2)
$y+3z+24=0$ (3)

And I tried to define y in terms of z, $y=3z-3z^2$ and insert in (3)
But I get the wrong answer after solving the new equation.

$(4,-36)$ and $(-2,-18)$

Would anyone mind to explain how to get to the right answer?

Thanks

2. Originally Posted by sebasto
I got this system of equations
$z-\lambda=0$ (1)
$3z^2+y-3\lambda=0$ (2)
$y+3z+24=0$ (3)

And I tried to define y in terms of z, $y=3z-3z^2$ and insert in (3)
But I get the wrong answer after solving the new equation.

$(4,-36)$ and $(-2,-18)$
If you substitute $y=3z-3z^2$ in (3), you should get $3z-3z^2+3z+24=0$, so that $3z^2 - 6z -24 = 0$, with solutions z = 4, –2. The given answer appears to list the coordinates in anti-alphabetical order, (z,y) rather than (y,z)!