Maximal volume of a Cuboid inscribed in an Ellipsoid

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SUMMARY

The discussion focuses on determining the maximal volume of a cuboid inscribed within the upper half of an ellipsoid defined by the equation D={(x,y,z): x^2/a^2 + y^2/b^2 + z^2/c^2 <=1, z>=0}. The author employs Lagrange's multipliers to derive the volume function v(x,y,z) = 4xyz and the constraint d(x,y,z) = x^2/a^2 + y^2/b^2 + z^2/c^2 = 1. The resulting dimensions of the cuboid are x = a/√3, y = b/√3, and z = c/√3, leading to a maximal volume of Vmax = 4/9 √3 abc. The solution is confirmed as correct by the author.

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  • Familiarity with ellipsoidal geometry
  • Knowledge of multivariable calculus
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I have an important paper to submit and I have a feeling I didn't solve the following correctly.


I have to find the maximal volume of a Cuboid inscribed inside half of the Ellipsoid
D={(x,y,z): x^2/a^2 + y^2/b^2 + z^2/c^2 <=1, z>=0 }

So I decided to use Lagrange's multipliers.

That's what I got:

v(x,y,z) = 4xyz
d(x,y,z) = x^2/a^2 + y^2/b^2 + z^2/c^2 = 1

∇v = λ ∇d
2yz = λ x/a^2
2xz = λ y/b^2
2xy = λ z/c^2

⇒ y/x = b/a and z/y = c/b
y^2/b^2 = 1/3 ... plug into d
y = b/√3
x = a/√3
z = c/√3
Vmax = 4/9 √3 abc

Answer: Vmax = 4/9 √3 abc

Is this OK?
 
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Looks ok to me.
 

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