Constrained optimization troubles

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SUMMARY

The discussion focuses on finding the largest volume of a regular parallelepiped inscribed in the ellipsoid defined by the equation x²/a² + y²/b² + z²/c² = 1 using the Lagrangian method. The user attempts to minimize the function f = xyz under the constraint provided, leading to a system of equations derived from the Lagrange multiplier method. The resulting equations indicate that if any of x, y, or z equals zero, the volume is minimized to zero, prompting the need to eliminate the Lagrange multiplier λ to find non-zero solutions. The user concludes that substituting y and z in terms of x allows for solving the ellipsoid constraint effectively.

PREREQUISITES
  • Understanding of Lagrangian optimization techniques
  • Familiarity with ellipsoidal geometry and equations
  • Knowledge of solving systems of equations
  • Basic calculus, particularly differentiation and optimization
NEXT STEPS
  • Study advanced applications of the Lagrange multiplier method in optimization problems
  • Explore geometric properties of ellipsoids and their applications in optimization
  • Learn about numerical methods for solving nonlinear equations
  • Investigate alternative optimization techniques such as gradient descent
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Mathematicians, optimization specialists, and students studying advanced calculus or applied mathematics who are interested in constrained optimization problems and geometric applications.

Eric Worceste
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I'm trying to find the regular parallelepiped with sides parallel to the coordinate axis inscribed in the ellipsoid x[2]/a[2] + y[2]/b[2] + z[2]/c[2] = 1 that has the largest volume. I've been trying the Lagrangian method: minimize f = (x)(y)(z), subject to the constraint (x[2]/a[2] + y[2]/b[2] + z[2]/c[2] - 1 = 0.

My new function F = (x)(y)(z) + L(x[2]/a[2] + y[2]/b[2] + z[2]/c[2] - 1) however leads to zeroes after differentiation and the linear system is solved. I'm stuck . . .
 
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The linear system is solved? What's bad about that? If you mean x= 0, y= 0, z= 0 are solutions, that's a solution but is not the only solution.
What you get, using the LaGrange multiplier method, is:
[itex]yz=2\lambda x/a[/itex]
[itex]xz= 2\lamda y/b[/itex]
[itex]xy= 2\lambda z/c[/itex]

Since anyone of those being 0 will give volume 0, that's obviously the minimum. Assuming none are 0, we can eliminate [itex]\lambda[/itex] by dividing one equation by another. For example, dividing the second equation by the first gives x/y= ay/bx which, in turn gives [itex]y^2= b^2x/a^2[/itex] so (we can assume x and y are positive here) y= bx/a. Similarly, dividing the second equation by the third results in z= cx/a. replace y and z in [itex]x^2/a^2+ y^2/b^2+ z^2/c^2= 1[/itex] and solve for x.
 

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