Question

Use lagrange multipliers to find three positive numbers whose sum is 120 and whose product is maximum. (enter your answers as a comma-separated list.) (40,40,40) incorrect: your answer is incorrect.

Answer 1

We're maximizing $xyz$ subject to $x+y+z=120$. We have Lagrangian

$L(x,y,z,\lambda)=xyz+\lambda(x+y+z-120)$

with partial derivatives (set to 0)

$L_x=yz+\lambda=0\implies yz=-\lambda$
$L_y=xz+\lambda=0\implies xz=-\lambda$
$L_z=xy+\lambda=0\implies xy=-\lambda$
$L_\lambda=x+y+z-120=0$

$yz=xz\implies z(x-y)=0\implies z=0\text{ or }x=y$
$xz=xy\implies x(y-z)=0\implies x=0\text{ or }y=z$
$yz=xy\implies y(x-z)=0\implies y=0\text{ or }x=z$

Since $x,y,z>0$, we arrive at $x=y=z$, which means

$x+y+z=3x=120\implies x=y=z=40$

and we get a maximum value of $40^3=64000$ for the product.