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+7y+11z=77?

Question

2926 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-07T15:06:59

the volume is # 5929/27 approx 219.6 " cubic units"#

you're optimising volume function

#f(x,y,z) = xyz#

with constraint #g(x,y,z) = x + 7y + 11z = 77#

using the Lagrange Multiplier ie #nabla f = lambda nabla g#

we get

#[(yz),(xz),(xy)] = lambda [(1),(7),(11)]#

comparing #lambda#'s and from rows 1 and 2

#(yz)/1 = (xz)/7 implies x = 7y#

from rows 2 and 3

#(xz)/7 = (xy)/11 implies z = (7y)/11#

if we park those back into the constraint

#7y + 7y + 11( (7y)/11 ) = 77#

#implies y = 77/21#

so #x = 77/3, z = 7/3#

So the volume is #77/3 * 77/21 * 7/3 = 5929/27 approx 219.6 " cubic units"#

in terms of scoping for a reality check, the total volume under that plane in the first octant is

#\int_(x=0)^(77) quad \int_(y=0)^( 11 - x/7) quad (77-x-7y)/11 quad dy \ dx approx 988 " cubic units"#