 # 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 Alyce Wilkinson 2021-02-25 Answered

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 $x+3y+5z=15$

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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,
$V=f\left(x,y\right)=xyz$
The vertex get lies in the plane as below,
$x+3y+5z=15$
$5z=15–x–3y$
$z=15–x–3\frac{y}{5}$
The volume is determined as below,
$V=f\left(x,y\right)=\frac{1}{5}xy\left(15-x-3y\right)=3xy-\frac{{x}^{2}y}{5}-\frac{3x{y}^{2}}{5}$
The Volume will be maximum if ${f}_{x}={f}_{y}=0$
As $f\left(x\right)=3y-\frac{2}{5}xy-\frac{3}{5}{y}^{2}=\frac{1}{5}y\left(15-2x-3y\right)=0$
$y=0$, $y=\frac{15-2x}{3}$ (1)
And
${f}_{y}=3x-\frac{{x}^{2}}{5}-\frac{6xy}{5}=\frac{1}{5}x\left(15-x-6y\right)=0$
$x=0$ and substituting $y=0$
$x=15$
At $y=\frac{15-2x}{3}$
${f}_{y}=0$
$\frac{1}{5}x\left(15-x-6\cdot \frac{15-2x}{3}\right)=0$
$x=0,15-x-30+4x=0$
$3x=15$
$x=5$
$y=\frac{15-2\cdot 0}{3}=\frac{15}{3}$
At $x=0$
$y=5$
And at $x=5$, $y=\frac{5}{3}$
The Critical point are $\left(0,0\right),\left(15,0\right),\left(0,5\right),\left(5,\frac{5}{3}\right)$ and z
At $\left(5,\frac{5}{3}\right)isz=\frac{15-x-3y}{5}$
$z=\frac{15-5-5}{5}$
$z=1$
The Volume of the longest rectangular box is determined as below,
$5\cdot \frac{5}{3}=\frac{25}{3}$
Hence, the volume is $\frac{25}{3}$