Question

Find three positive integers x, y, and z that satisfy the given conditions. the product is 64, and the sum is a minimum.

Answer 1

Minimize $x+y+z$ subject to $xyz=64$. Using Lagrange multipliers, we can take the Lagrangian

$L(x,y,z,\lambda)=x+y+z+\lambda(xyz-64)$

which has partial derivatives (set to 0)

$\begin{cases}L_x=1+\lambda yz=0\\L_y=1+\lambda xz=0\\L_z=1+\lambda xy=0\\L_\lambda=xyz-64=0\implies xyz=64\end{cases}$

From the first three equations, pick any two and subtract them from one another. You'll arrive at the following symmetric relations:

$\begin{cases}(1+\lambda yz)-(1+\lambda xz)=0\implies \lambda(y-x)z=0\\\lambda(z-y)x=0\\\lambda(z-x)y=0\end{cases}$

We assume $\lambda\neq0$, and we know that $xyz=64$ so that we can omit the possibilities of $x=0$, $y=0$, and $z=0$. This leaves us with $x=y=z$, and so

$xyz=64=x^3=y^3=z^3\implies x=y=z=4$.