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 + 5z = 24?

Answer 1

#V=64/5# (#units^3#)

The volume of a rectangular box is given by the formula #V=xyz# (equivalent to #V=lwh#).
We want to maximize #V# given the constraint #x+8y+5z=24#.
#gradf=< f_x,f_y,f_z > => < yz, xz, xy >#
#gradg= < g_x,g_y,g_z > => < 1, 8, 5 >#
This gives #< yz, xz, xy > = lambda< 1, 8, 5 >#

Now we set the respective components equal:

There are no general rules for solving systems of equations, so some ingenuity may be required at times. Common strategies involve solving for the variables #x,y,z# or #lambda# to begin. Which you use will depend on the particular system you are dealing with. In this case, I would solve for #lambda#.
#lambda=yz#
#lambda=(xz)/8#
#lambda=(xy)/5#

We can set the equations equal:

#yz=(xz)/8=(xy)/5#

From equations 1 and 2:

#yz=(xz)/8#
#=>8y=x#

From equations 1 and 3:

#yz=(xy)/5#
#=>5z=x#

We can substitute these (convenient) values back into the constraint:

#x+x+x=24#
#=>3x=24#
#=>x=8#
This gives that #z=8/5# and #y=1#.

You may check these values:

#8+8(1)+5(8/5)=24#
#8+8+8=24#
#24=24#
Therefore, the volume is #64/5# (#units^3#)

A formal explanation (1):

Method of Lagrange Multipliers

To find the maximum and minimum values of #f(x,y,z)# subject to the constraint #g(x,y,z)=k# [assuming that these extreme values exist and #gradg!=0# on the surface #g(x,y,z)=k#]:
(a) Find all values of #x,y,z# and #lambda# such that
#gradf(x,y,z)=lambda gradg(x,y,z)# #g(x,y,z)=k#

and

(b) Evaluate #f# at all points #(x,y,z)# that result from step (a). The largest of these values is the maximum value of #f#; the smallest is the minimum value of #f#.
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Answer 2

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 + 5z = 24), follow these steps:

  1. Define the objective function: (V = xyz), where (x), (y), and (z) are the dimensions of the rectangular box.

  2. Formulate the constraint equation: (g(x, y, z) = x + 8y + 5z - 24 = 0), representing the given plane.

  3. Set up the Lagrangian function: (L(x, y, z, \lambda) = xyz + \lambda(x + 8y + 5z - 24)), where (\lambda) is the Lagrange multiplier.

  4. Find the partial derivatives of the Lagrangian function with respect to (x), (y), (z), and (\lambda), and set them equal to zero:

    [ \begin{align*} \frac{\partial L}{\partial x} &= yz + \lambda = 0 \ \frac{\partial L}{\partial y} &= xz + 8\lambda = 0 \ \frac{\partial L}{\partial z} &= xy + 5\lambda = 0 \ \frac{\partial L}{\partial \lambda} &= x + 8y + 5z - 24 = 0 \end{align*} ]

  5. Solve this system of equations to find the critical points.

  6. Check the critical points to determine which one yields the maximum volume.

  7. Substitute the values of (x), (y), and (z) obtained from the critical point into the objective function (V = xyz) to find the maximum volume of the rectangular box.

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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