Math Problem Statement

Find the maximum volume of a rectangular parallelepiped that can be inscribed in a sphere using Lagrange's multipliers method.

Solution

To solve this problem, we aim to maximize the volume of a rectangular parallelepiped inscribed in a sphere using the method of Lagrange multipliers. Here’s the step-by-step solution:


Step 1: Problem setup

  1. Let the rectangular parallelepiped have dimensions x,y,zx, y, z, so its volume is: V=xyzV = x \cdot y \cdot z
  2. Since the parallelepiped is inscribed in a sphere, the constraint is that the distance from the origin to any corner of the parallelepiped lies on the sphere. This implies: g(x,y,z)=x2+y2+z2R2=0g(x, y, z) = x^2 + y^2 + z^2 - R^2 = 0 where RR is the radius of the sphere.

Step 2: Lagrange multipliers

To maximize V(x,y,z)V(x, y, z) subject to g(x,y,z)=0g(x, y, z) = 0, we solve: V=λg\nabla V = \lambda \nabla g where V=(yz,xz,xy)\nabla V = (yz, xz, xy) and g=(2x,2y,2z)\nabla g = (2x, 2y, 2z).

This leads to the system of equations:

  1. yz=λ(2x)yz = \lambda (2x)
  2. xz=λ(2y)xz = \lambda (2y)
  3. xy=λ(2z)xy = \lambda (2z)
  4. x2+y2+z2=R2x^2 + y^2 + z^2 = R^2 (constraint)

Step 3: Solve for λ\lambda

From the first three equations: λ=yz2x=xz2y=xy2z\lambda = \frac{yz}{2x} = \frac{xz}{2y} = \frac{xy}{2z} Equating any two pairs:

  1. From yz2x=xz2y\frac{yz}{2x} = \frac{xz}{2y}, we get y2=x2y^2 = x^2, so y=±xy = \pm x.
  2. From yz2x=xy2z\frac{yz}{2x} = \frac{xy}{2z}, we get z2=x2z^2 = x^2, so z=±xz = \pm x.

Step 4: Substituting into the constraint

Assume x=y=zx = y = z (symmetry leads to a positive maximum): x2+x2+x2=R2    3x2=R2    x=R3x^2 + x^2 + x^2 = R^2 \implies 3x^2 = R^2 \implies x = \frac{R}{\sqrt{3}}

Thus: x=y=z=R3x = y = z = \frac{R}{\sqrt{3}}


Step 5: Maximum volume

The maximum volume is: V=xyz=(R3)3=R333V = x \cdot y \cdot z = \left( \frac{R}{\sqrt{3}} \right)^3 = \frac{R^3}{3\sqrt{3}}


Final Answer:

The maximum volume of the rectangular parallelepiped is: V=R333V = \frac{R^3}{3\sqrt{3}}


Would you like further clarification on any step? Here are related questions for deeper exploration:

  1. Why does symmetry ensure x=y=zx = y = z for maximum volume in this case?
  2. How does the method of Lagrange multipliers guarantee optimal solutions under constraints?
  3. Can this method be adapted to maximize the volume of shapes inscribed in ellipsoids?
  4. How does the result change if the parallelepiped has specific fixed aspect ratios (e.g., x:y:zx : y : z)?
  5. What are practical applications of maximizing the volume of an object under constraints?

Tip: Always confirm your constraint and symmetry assumptions when solving geometry-based optimization problems!

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Math Problem Analysis

Mathematical Concepts

Optimization
Multivariable Calculus
Geometry

Formulas

Volume of rectangular parallelepiped: V = x * y * z
Constraint for a sphere: x^2 + y^2 + z^2 = R^2

Theorems

Lagrange Multipliers

Suitable Grade Level

Undergraduate Mathematics