Question

G use lagrange multipliers to find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the given plane.x + 9y + 8z = 27

Answer 1

This basically comes down to maximizing $xyz$ subject to $x+9y+8z=27$ and enforcing $x,y.z>0$. We have the Lagrangian

$L(x,y,z,\lambda)=xyz+\lambda(x+9y+8z-27)$

with partial derivatives (set equal to 0 to find critical points)

$\begin{cases}L_x=yz+\lambda=0\\L_y=xz+9\lambda=0\\L_z=xy+8\lambda=0\\L_\lambda=x+9y+8z-27=0\end{cases}$

Solving the first equation for $\lambda$ gives $\lambda=-yz$. Substituting this into the next two equations, we have

$xz-9yz=0\implies z(x-9y)=0\implies x=9y$
$xy-8yz=0\implies y(x-8z)=0\implies x=8z$

Now

$x+9y+8z=27\implies x+x+x=3x=27\implies x=9$
$x=9y\implies y=1$
$x=8z\implies z=\dfrac98$

So the vertex of the cuboid in the given plane that maximizes the cuboids volume is $(x,y,z)=\left(9,1,\dfrac98\right)$, giving a volume of $\dfrac{81}8$.