Answer
$\dfrac{4}{3} $
Work Step by Step
Given: Equation of plane : $x+2y+3z=6$
Re-write the equation of plane as: $x=6-2y-3z$ ...(1)
Need to apply Lagrange Multipliers Method to determine the maximum volume of a rectangular box.
we have $\nabla f=\lambda \nabla g$
The volume of a rectangular box is $V=xyz$
From equation (1), we have $V=6yz-2y^2z-3yz^2$
Also, $V_y=6z-4yz-3z^2, V_z=6y-2y^2-6yz$
Simplify to get the values of $y$ and $z$.
we have $y=1,z=\dfrac{2}{3}$
and $x=6-2y-3z=6-2(1)-3(\dfrac{2}{3})=2$
Thus, the volume of a rectangular box is
$V=xyz=(2)(1)(\dfrac{2}{3})$
$V=\dfrac{4}{3} $
