Determine the volume of the largest box in the first octant with three in the coordinate planes, one vertex at the origin, and its opposite vertex in the plane x+3y+5z=15

The volume of the largest rectangular box is to be determined based on the first octant with three faces in the coordinate planes, and one vertex in the plane $x+3y+5z=15$.
The volume of the rectangular box that lies in the first octant based on the three faces that lies in the coordinate plane as follows,
${\displaystyle V=f(x,y)=xyz}$
The vertex get lies in the plane as below,
$x+3y+5z=15$
$5z=15–x–3y$
${\displaystyle z=15–x–3\frac{y}{5}}$
The volume is determined as below,
${\displaystyle V=f(x,y)=\frac{1}{5}xy(15-x-3y)=3xy-\frac{x^{2}y}{5}-\frac{3x{y}^2}{5}}$
The Volume will be maximum if ${\displaystyle f_x=f_y=0}$
As ${\displaystyle f\left(x\right)=3y-\frac{2}{5}xy-\frac{3}{5}{y}^2=\frac{1}{5}y(15-2x-3y)=0}$
$y=0$, ${\displaystyle y=\frac{15-2x}{3}}$ (1)
And
${\displaystyle f_y=3x-\frac{x^2}{5}-\frac{6xy}{5}=\frac{1}{5}x(15-x-6y)=0}$
$x=0$ and substituting $y=0$
$x=15$
At ${\displaystyle y=\frac{15-2x}{3}}$
${\displaystyle f_y=0}$
${\displaystyle \frac{1}{5}x(15-x-6\cdot \frac{15-2x}{3})=0}$
$x=0,15-x-30+4x=0$
$3x=15$
$x=5$
${\displaystyle y=\frac{15-2\cdot 0}{3}=\frac{15}{3}}$
At $x=0$
$y=5$
And at $x=5$, ${\displaystyle y=\frac{5}{3}}$
The Critical point are ${\displaystyle (0,0),(15,0),(0,5),(5,\frac{5}{3})}$ and z
At ${\displaystyle (5,\frac{5}{3})isz=\frac{15-x-3y}{5}}$
$z=\frac{15-5-5}{5}$
$z=1$
The Volume of the longest rectangular box is determined as below,
${\displaystyle 5\cdot \frac{5}{3}=\frac{25}{3}}$
Hence, the volume is ${\displaystyle \frac{25}{3}}$