Maximizing the Volume of a Rectangular Parallelepiped Inscribed in an Ellipsoid Using Lagrange Multipliers

Optimizing the volume of a geometric shape inscribed within a specific boundary can be a fascinating challenge in multivariable calculus. One such problem involves finding the volume of the largest rectangular parallelepiped that can be inscribed in an ellipsoid defined by the equation (frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2} 1). This problem can be elegantly solved using the method of Lagrange Multipliers, a powerful tool in multivariable optimization.

Understanding the Problem

The goal is to determine the dimensions of a rectangular parallelepiped (a general four-sided prism with rectangular faces) that can be inscribed within the ellipsoid and has the maximum possible volume. A rectangular parallelepiped inscribed in an ellipsoid means that its vertices touch the ellipsoid's surface. The key is to maximize the volume (V 8xyz) (since the parallelepiped is centered at the origin and its dimensions extend to (2x), (2y), and (2z)) while satisfying the constraint defined by the ellipsoid equation.

Setting Up the Lagrange Multipliers

The volume (V 8xyz) serves as the function to be maximized, and the constraint provided by the ellipsoid is given by (frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2} 1). We introduce a Lagrange multiplier (lambda) and form the Lagrangian function:

[mathcal{L}(x, y, z, lambda) 8xyz lambda left(1 - left(frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2}right)right)]

Next, we find the partial derivatives of (mathcal{L}) with respect to (x), (y), (z), and (lambda), and set them equal to zero:

[frac{partial mathcal{L}}{partial x} 8yz - lambda frac{2x}{a^2} 0]

[frac{partial mathcal{L}}{partial y} 8xz - lambda frac{2y}{b^2} 0]

[frac{partial mathcal{L}}{partial z} 8xy - lambda frac{2z}{c^2} 0]

[frac{partial mathcal{L}}{partial lambda} 1 - left(frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2}right) 0]

Solving the System of Equations

From the first three equations, we can isolate (lambda):

[lambda frac{4a^2yz}{x} frac{4b^2xz}{y} frac{4c^2xy}{z}]

From (lambda frac{4a^2yz}{x}), we get (frac{a^2yz}{x} frac{b^2xz}{y}). By solving for (lambda), we obtain:

[lambda frac{4yz}{1} frac{4xz}{1} frac{4xy}{1}]

Equating (frac{a^2yz}{x} frac{b^2xz}{y}) yields (frac{a^2}{x} frac{b^2}{y}), and similarly for the other pairs. This implies:

[frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2}]

Substituting these into the constraint equation (frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2} 1) leads to:

[frac{x^2}{a^2} frac{y^2}{b^2} frac{z^2}{c^2} frac{1}{3}]

Therefore, (x pm frac{a}{sqrt{3}}), (y pm frac{b}{sqrt{3}}), and (z pm frac{c}{sqrt{3}}). The volume of the largest rectangular parallelepiped is then:

[V 8xyz 8 left(frac{a}{sqrt{3}}right)left(frac{b}{sqrt{3}}right)left(frac{c}{sqrt{3}}right) frac{8abc}{3sqrt{3}}]

Conclusion

Using the method of Lagrange Multipliers, we successfully determined the dimensions of the rectangular parallelepiped that maximize its volume within the ellipsoid. The key steps involved setting up the volume function, incorporating the constraint, and solving the resulting system of equations. This problem not only demonstrates the power of Lagrange Multipliers in solving real-world optimization problems but also provides a deeper understanding of ellipsoidal geometry and its applications in various fields of mathematics and engineering.