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#?
This box volume is described as
subject to
The lagrangian is
Here
Now the stationary points are given by the solutions of
#{(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):}#
#(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)#
The found volume is
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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:

Define the variables:
 Let (x, y, z) be the lengths of the sides of the rectangular box.

Define the function to optimize:
 The volume of the rectangular box is given by (V = xyz).

Define the constraint equation:
 The constraint is the equation of the plane (x + 8y + 7z = 24).

Formulate the Lagrangian:
 The Lagrangian function is: [ L(x, y, z, \lambda) = xyz  \lambda(x + 8y + 7z  24) ]

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 ]

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.


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

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.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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