Let the 2 parts be x,y

So, 3rd part is 24-x-y

Let the function to be maximized be f

$f(x,y)=x^2\times y^3 \times (24-x-y)$

$\log f=2 \log x + 3 \log y + \log (24 -x-y)$

Partially differently w.r.t x,

$\dfrac {1}{f}f_x = \dfrac {2}{x} - \dfrac {1}{24-x-y} \\ f_x = d \left [ \dfrac {2}{x}- \dfrac {1}{24-x-y} \right ]$

Partially different w.r.t y,

$\dfrac {1}{f}f_y =\dfrac {3} {y}- \dfrac {1}{24-x-y} $

$f_y =f \left [ \dfrac {3}{y}- \dfrac {1}{24-x-y} \right ] $

$f_{xx}=f \left [ - \dfrac {2}{x_2}- \dfrac {1}{(24-x-y)^2} \right ]+ f_x \left [ \dfrac {2}{x} - \dfrac {1}{24-x-y} \right ] $

$f_{xy} = f_y \left [ \dfrac {3}{y} - \dfrac {1}{24-x-y}\right ] - f \dfrac {1}{(24-x-y)^2} $

$f_{yy}=f_y \left [ \dfrac {3}{y}- \dfrac {1}{24-x-y} \right ]+ f\left [ - \dfrac {3}{y^2} - \dfrac {1}{(24-x-y^2)} \right ]$

$f_x=0 \ and \ f_y=0 \ gives \ extreme \ values \ of \ f \ (f\ne 0)$

$\dfrac {2}{x}- \dfrac {1}{24-x-y}=0 \ and \ \dfrac {3}{y} - \dfrac {1}{24-x-y}=0 $

$48-2x-2y-x=0, \ 72-3x-3y-y=0 $

$3x+2y=48, \ 3x+4y=72$

$48-2y=72-4y $

$24=2y$

$y=12$

$from , \ 3x+2y=48,$

$3x+24=48$

$x=8.$

The three parts are 4,8,12.

$f=4(64) (1728)$

$Also, \ f_{xx} = f \left ( -\dfrac {1}{32}- \dfrac {1}{16} \right )= - \dfrac {3f}{32} $

$f_{xy}= -f \left ( \dfrac {1}{16} \right )= -\dfrac {f}{16} $

$f_{yy}= f \left ( - \dfrac {1}{48}- \dfrac {1}{16} \right )= - \dfrac {f}{12} $

(Hence, \ f_{xx}f_{yy}> f^2_{xy} )

(Also , \ f_{xx}<0 )

$\therefore f(x,y)\ is \ maximum \ at \ x=8, \ y=12.$