Answer: The volume of largest rectangular box is 4.5 units.
Step-by-step explanation:
Since we have given that
Volume = [tex]xyz[/tex]
with subject to [tex]x+2y+3z=9[/tex]
So, let [tex]z=\dfrac{9-x-2y}{3}[/tex]
So, Volume becomes,
[tex]V=xyz\\\\V=xy(\dfrac{9-x-2y}{3})\\\\V=\dfrac{9xy-x^2y-2xy^2}{3}[/tex]
Partially derivative wrt x and y we get that
[tex]9-2x-2y=0\implies 2x+2y=9\\\\and\\\\9-x-4y=0\implies x+4y=9[/tex]
By solving these two equations, we get that
[tex]x=3,y=\dfrac{3}{2}[/tex]
So, [tex]z=\dfrac{9-x-2y}{3}=\dfrac{9-3-3}{3}=\dfrac{3}{3}=1[/tex]
So, Volume of largest rectangular box would be
[tex]xyz=3\times \dfrac{3}{2}\times 1=\dfrac{9}{2}=4.5[/tex]
Hence, the volume of largest rectangular box is 4.5 units.
