Answer
The three positive numbers are $\left( {x,y,z} \right) = \left( {50,50,50} \right)$, which has the largest possible product $125000$.
Work Step by Step
Let the three positive numbers be $x$, $y$ and $z$. It is required that $x+y+z=150$.
Write the product of the three: $f=x y z$.
Using $x+y+z=150$, we can write $f$ as a function of $x$ and $y$:
$f\left( {x,y} \right) = xy\left( {150 - x - y} \right) = 150xy - {x^2}y - x{y^2}$
The partial derivatives are
${f_x} = 150y - 2xy - {y^2}$, ${\ \ }$ ${f_y} = 150x - {x^2} - 2xy$
${f_{xx}} = - 2y$, ${\ \ }$ ${f_{yy}} = - 2x$, ${\ \ }$ ${f_{xy}} = 150 - 2x - 2y$
We find the critical points of $f$ by solving ${f_x} = 0$ and ${f_y} = 0$:
$150y - 2xy - {y^2} = 0$, ${\ \ }$ $150x - {x^2} - 2xy = 0$
$y\left( {150 - 2x - y} \right) = 0$, ${\ \ }$ $x\left( {150 - x - 2y} \right) = 0$
From the first equation, we obtain $y=0$. Substituting it in the second equation gives $x=150$.
From the second equation, we obtain $x=0$. Substituting it in the first equation gives $y=150$.
From the equations $150-2x-y=0$ and $150-x-2y=0$, we obtain the solution $x=50$ and $y=50$.
So, the critical points are $\left( {0,0} \right)$, $\left( {150,0} \right)$, $\left( {0,150} \right)$ and $\left( {50,50} \right)$. The corresponding extreme values are $f\left( {0,0} \right) = 0$, $f\left( {150,0} \right) = 0$, $f\left( {0,150} \right) = 0$ and $f\left( {50,50} \right) = 125000$.
Since ${f_{xx}} = - 100 < 0$ and the discriminant $D = {f_{xx}}{f_{yy}} - {f_{xy}}^2 = 7500 > 0$ at $\left( {50,50} \right)$, by Theorem 2, $f\left( {50,50} \right) = 125000$ is a local maximum. Thus, the largest possible product of the three numbers is $125000$.
Using the equation $x+y+z=150$, we obtain the three numbers $\left( {x,y,z} \right) = \left( {50,50,50} \right)$.
