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.
