# Thread: Lagrange multipliers with 2 constraints

1. ## Lagrange multipliers with 2 constraints

I haven't done this for a long time so i'm bit stuck on this
Max. $\displaystyle V=xyz$ for $\displaystyle x,y,z \geq 0$ with constraints
$\displaystyle xy + yz + zx = 1$
$\displaystyle x + y + z = 3$

I found:
$\displaystyle h = xyz + \lambda(xy+yz+zx-1) + \mu(x+y+z-3)$
$\displaystyle \frac{\partial}{\partial x}=yz + \lambda(y+z) + \mu$
$\displaystyle \frac{\partial}{\partial y}=xz + \lambda(y+x) + \mu$
$\displaystyle \frac{\partial}{\partial z}=xy + \lambda(y+x) + \mu$
$\displaystyle \frac{\partial}{\partial \lambda}=xy +yz + zx -1$
$\displaystyle \frac{\partial}{\partial \mu}=x + y + z -3$

Think this is correct step up to this point, but when I try to rearrange above equation to find $\displaystyle \lambda and \mu$ I couldn't do it.

2. Originally Posted by kleenex
$\displaystyle \frac{\partial}{\partial x}=yz + \lambda(y+z) + \mu$
$\displaystyle \frac{\partial}{\partial y}=xz + \lambda({\color{red}z}+x) + \mu$
$\displaystyle \frac{\partial}{\partial z}=xy + \lambda(y+x) + \mu$
Putting these equal to 0, we have
$\displaystyle yz + \lambda(y+z) + \mu = 0$ ..... (1)
$\displaystyle xz + \lambda(z+x) + \mu = 0$ ..... (2)
$\displaystyle xy + \lambda(y+x) + \mu = 0$ ..... (3)

Subtract (2) from (1): $\displaystyle (z + \lambda)(y-x) = 0$. Therefore either $\displaystyle x=y$ or $\displaystyle \lambda = -z$. Similarly
either $\displaystyle y=z$ or $\displaystyle \lambda = -x$; and
either $\displaystyle z=x$ or $\displaystyle \lambda = -y$.

From those equations it clearly follows that x, y and z cannot all be different. On the other hand, they cannot all be the same, because then the equation x+y+z=3 would mean that they are all equal to 1, which would contradict the equation yz+zx+xy=1.

Thus two of x,y and z are equal, and the third is different, say z=y≠x. Then the equations yz+zx+xy=1 and x+y+z=3 become $\displaystyle 2xy+y^2=1$, $\displaystyle x+2y=3$. Substitute the value of x from the second of those equations into the first to get a quadratic for y. The numbers don't come out particularly pleasantly, but I get the maximum value of V as $\displaystyle -1+{\textstyle\frac49}\sqrt6\approx0.0866$.