How do you find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x + 2y + 3z = 5?

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Answer 1 · 2017-03-07T14:14:11

Volume of largest rectangular box is #125/162#

The volume of the rectangular box in the first octant with three faces in the coordinate planes will be #V=f(x,y)=xyz#.

As the vertex lies in the plane #x+2y+3z=5#, #z=(5-x-2y)/3# and volume is

#V=f(x,y)=1/3xy(5-x-2y)=5/3xy-1/3x^2y-2/3xy^2#

Volume will be maximum if #f_x=f_y=0#

As #f_x=5/3y-2/3xy-2/3y^2=1/3y(5-2x-2y)=0# which implies

#y=0#, #y=5/2-x# ...............(1)

and #f_y=5/3x-1/3x^2-4/3xy=1/3x(5-x-4y)=0# ...............(2)

Substituting #y=0# in (2)

#1/3x(5-x)=0=>#, #x=0#, #x=5#

and at #y=5/2-x#

#1/3x(5-x-10+4x)=0# i.e.

#x(-5+3x)=0=>#, #x=0#, #x=5/3#

At #x=0# #y=5/2# and at #x=5/3# #y=5/6#

So critical points are #(0,0)#, #(5,0)#, #(0,5/2)# and #(5/3,5/6)#

and #z# at #(5/3,5/6)# is #z=(5-5/3-2xx5/6)/3=5/9#

and volume of largest rectangular box is #5/3xx5/6xx5/9=125/162#