Question

Find the volume of the largest rectangular box with edges parallel to the axes that can be inscribed in the ellipsoid

(x^2/4) + (y^2/64) + (z^2/49) = 1
Hint: By symmetry, you can restrict your attention to the first octant (where x,y,z=0), and assume your volume has the form V=8xyz. Then arguing by symmetry, you need only look for points which achieve the maximum which lie in the first octant.
What is the maximum volume?

Answer 1

The maximum volume of the rectangular box inscribed in the given ellipsoid
`\( \left((x^2)/(4)\right) + \left((y^2)/(64)\right) + \left((z^2)/(49)\right) = 1 \) is \( (64)/(3) \)` cubic units.

Step-by-step explanation:

To find the maximum volume of a rectangular box inscribed in the ellipsoid, we consider the first octant for simplicity, where ( x, y, z ) are positive. Employing symmetry, the maximum volume occurs when the box touches the ellipsoid at points with maximum coordinates ( (x, y, z) ) within the first octant.

Utilizing Lagrange multipliers or similar methods, it can be deduced that the maximum volume is attained when
`\( (x)/(2) = (y)/(8) = (z)/(7) \).` Thus, the dimensions of the box become \( x = 2t, y = 8t, z = 7t \), and substituting these into
`\( \left((x^2)/(4)\right) + \left((y^2)/(64)\right) + \left((z^2)/(49)\right) = 1 \) yields \( t = (1)/(8) \). C`onsequently,
`\( x = (1)/(4), y = 1, z = (7)/(8) \).`

Finally, the volume
`\( V = 8xyz \)`equals
`\( (64)/(3) \)` cubic units, representing the maximum volume achievable for the inscribed rectangular box.

"".

Answer 2

The maximum volume of the largest rectangular box that can be inscribed in the given ellipsoid is
`\( 40√(2) \)` cubic units.

To find the volume of the largest rectangular box that can be inscribed in the ellipsoid:

`(x^2)/(4) + (y^2)/(25) + (z^2)/(16) = 1`

we'll use the method of Lagrange multipliers, which is a strategy for finding the local maxima and minima of a function subject to equality constraints.

First, let's define our functions. The volume
`\( V \)` of the rectangular box is given by
`\( V = xyz \)`. Since we are considering the first octant where
`\( x, y, z \geq 0 \)`, and due to the symmetry of the problem, the volume of the inscribed box is
`\( V = 8xyz \)`.

The constraint is given by the equation of the ellipsoid:

`g(x, y, z) = (x^2)/(4) + (y^2)/(25) + (z^2)/(16) - 1 = 0`

The method of Lagrange multipliers tells us to find the stationary points of the Lagrange function
`\( L(x, y, z, \lambda) = 8xyz - \lambda \left( (x^2)/(4) + (y^2)/(25) + (z^2)/(16) - 1 \right) \)`, where
`\( \lambda \)` is the Lagrange multiplier.

We need to solve the system of equations given by the partial derivatives of
`\( L \)` with respect to
`\( x, y, z, \)` and
`\( \lambda \)`:

1.
`\( (\partial L)/(\partial x) = 8yz - (\lambda x)/(2) = 0 \)`

2.
`\( (\partial L)/(\partial y) = 8xz - (\lambda y)/(25) = 0 \)`

3.
`\( (\partial L)/(\partial z) = 8xy - (\lambda z)/(8) = 0 \)`

4.
`\( (\partial L)/(\partial \lambda) = (x^2)/(4) + (y^2)/(25) + (z^2)/(16) - 1 = 0 \)`

Let's solve this system step by step.

Step 1: Solve for
`\( \lambda \)`

From equations 1, 2, and 3, we can express
`\( \lambda \)` in terms of
`\( x, y, \)` and
`\( z \)`:

1.
`\( \lambda = (16yz)/(x) \)` (from equation 1)

2.
`\( \lambda = (200xz)/(y) \)` (from equation 2)

3.
`\( \lambda = (64xy)/(z) \)` (from equation 3)

Step 2: Equate Expressions for
`\( \lambda \)`

Equate the expressions for
`\( \lambda \)` from above:

1.
`\( (16yz)/(x) = (200xz)/(y) \)`

2.
`\( (200xz)/(y) = (64xy)/(z) \)`

3.
`\( (64xy)/(z) = (16yz)/(x) \)`

Simplify these equations to find relationships between
`\( x, y, \)` and
`\( z \)`.

1.
`16y^2z^2 = 200x^2z^2`

2.
`\( 200x^2z^2 = 64x^2y^2 \)`

3.
`\( 64x^2y^2 = 16y^2z^2 \)`

From these, we can derive relationships between
`\( x, y, \)` and
`\( z \)`.

Once we have these relationships, we can substitute them into the ellipsoid equation
`\( (x^2)/(4) + (y^2)/(25) + (z^2)/(16) = 1 \)` to find specific values for
`\( x, y, \)` and
`\( z \)`.

The solution to the system of equations yields several points. Among these, we are interested in the points in the first octant where
`\( x, y, z \geq 0 \)`.

Step 3: Analyzing the Solutions

The solutions to the system of equations are:

1.
`\( x = 0, y = 0, z = \pm 4 \)`

2.
`\( x = 0, y = \pm 5, z = 0 \)`

3.
`\( x = \pm 2, y = 0, z = 0 \)`

4.
`\( x = \pm 1, y = \pm (5)/(√(2)), z = \pm 2 \)`

Among these solutions, we are interested in those that lie in the first octant, where
`\( x, y, z \geq 0 \)`. This is because the ellipsoid is symmetric in all octants, and the problem can be restricted to the first octant without loss of generality.

The only solution that satisfies this condition is
`\( x = 1, y = (5)/(√(2)), z = 2 \)`. All other solutions either have negative coordinates or result in a zero volume (since one or more of the dimensions
`\( x, y, z \)` are zero).

Calculating the Maximum Volume:

Now, we calculate the volume of the box using the formula
`\( V = 8xyz \)`. This factor of 8 accounts for the fact that we are considering only the first octant and the box is symmetric in all octants.

1.
`\( x = 1, y = (5)/(√(2)), z = 2 \)`

2.
`\( x = 1, y = (5)/(√(2)), z = 2 \)`

Both these points are actually the same and represent the dimensions of the box that maximizes the volume within the given ellipsoid.

Now, let's calculate the maximum volume using these dimensions. The volume
`\( V \)` is given by
`\( V = 8xyz \)`. Substituting
`\( x = 1 \), \( y = (5)/(√(2)) \)`, and
`\( z = 2 \)`, we get:

`$$V = 8 * 1 * (5)/(√(2)) * 2.$$`

Let's compute this value.

The maximum volume of the largest rectangular box that can be inscribed in the given ellipsoid is
`\( 40√(2) \)` cubic units.