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#?

Question

William Abdalla · Accepted Answer

#V = 64/7#

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.

Emily Ammerman · Answer

undefined To 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 plane $x + 8y + 7z = 24$, follow these steps:

1. Define the variables:
   - Let $x, y, z$ be the lengths of the sides of the rectangular box.

2. Define the function to optimize:
   - The volume of the rectangular box is given by $V = xyz$.

3. Define the constraint equation:
   - The constraint is the equation of the plane $x + 8y + 7z = 24$.

4. Formulate the Lagrangian:
   - The Lagrangian function is:
     $$ L(x, y, z, \lambda) = xyz - \lambda(x + 8y + 7z - 24) $$

5. Find the partial derivatives:
   $$ \frac{\partial L}{\partial x} = yz - \lambda = 0 $$
   $$ \frac{\partial L}{\partial y} = xz - 8\lambda = 0 $$
   $$ \frac{\partial L}{\partial z} = xy - 7\lambda = 0 $$
   $$ \frac{\partial L}{\partial \lambda} = x + 8y + 7z - 24 = 0 $$

6. Solve the system of equations:
   - From the first three equations:
     $$ yz = \lambda $$
     $$ xz = 8\lambda $$
     $$ xy = 7\lambda $$

- Eliminate $x, y, z$ from these equations to get:
     $$ x = \frac{8\lambda}{z} $$
     $$ y = \frac{7\lambda}{z} $$

- Substitute these into the constraint equation $x + 8y + 7z = 24$:
     $$ \frac{8\lambda}{z} + 8\cdot\frac{7\lambda}{z} + 7z = 24 $$
     $$ \frac{8\lambda + 56\lambda}{z} + 7z = 24 $$
     $$ \frac{64\lambda}{z} + 7z = 24 $$

- Now solve for $z$:
     $$ 64\lambda + 7z^2 = 24z $$
     $$ 7z^2 - 24z + 64\lambda = 0 $$

- This is a quadratic equation in $z$, solve for $z$ to get two possible values.

7. Once you have the values of $z$, substitute them back into the expressions for $x$ and $y$ to get the corresponding values.

8. Finally, calculate the volume $V = xyz$ for each of the possible solutions, and choose the one that gives the maximum volume.

This will give you the dimensions of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane $x + 8y + 7z = 24$, along with its volume.