How do you 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 + 8y + 7z = 24#?

1 Answer
Oct 26, 2016

#V = 64/7#

Explanation:

This box volume is described as

#maxV=xyz#

subject to

#x - s_1^2= 0#
#y-s_2^2= 0#
#z-s_3^2 = 0#
#x + 8y + 7z = 24#

We introduce the slack variables #s_1,s_2,s_3# to transform the inequality into equality restrictions.

The lagrangian is

#L(X,S,Lambda)=xyz+lambda_1(x-s_1^2)+lambda_2(y-s_2^2)+lambda_3(z-s_3^2)+lambda_4(x+8y+7z-24)#

Here

#X=(x,y,z)#
#S=(s_1,s_2,s_3)#
#Lambda=(lambda_1,lambda_2,lambda_3,lambda_4)#

Now the stationary points are given by the solutions of

#grad L(X,S,Lambda)= vec 0#

#{(lambda_1 + lambda_4 + y z=0), (lambda_2 + 8 lambda_4 + x z=0), (lambda_3 + 7 lambda_4 + x y=0), (-2 lambda_1 s_1=0), (-2 lambda_2 s_2=0), (-2 lambda_3 s_3=0), (-s_1^2 + x=0), (-s_2^2 + y=0), (-s_3^2 + z=0), (-24 + x + 8 y + 7 z=0):}#

Solving for #(X,S,Lambda)# we have a meaningful sample

#(x = 8, y = 1, z = 8/7, s_1 = -2 sqrt[2], s_2= -1, s_3= 2 sqrt[2/7], lambda_1= 0, lambda_2= 0, lambda_3= 0, lambda_4 = -8/7)#

we rejected stationary points with #x=0# or #y=0# or #z=0#
which implied on null volumes.

The shown solution has #s_1 ne 0, s_2 ne0, s_3 ne 0# so is an interior solution.

The #Lambda# found #lambda_1=lambda_2=lambda_3=0, lambda_4 ne 0# show the actuating and non actuacting restrictions. The #lambda_k = 0# for non actuating restrictions and #lambda_4 ne 0# for the actuating one.

The found volume is

#V = 64/7# volume units.