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I was working on Lagrange Multipliers but I want to find another method using Geometry..

From calculus book the question is, with the information given,

"Find the dimensions of the rectangular box of maximum volume that can be inscribed inside the sphere $x^2 +y^2 +z^2 = 4$"

I understand how to do it their way and I do not seek an optimization approach or a Lagrange Multiplier approach but a geometric one with shapes/triangles and algebra and reason behind the process.

The answer is that the volume consists of all like edges, $\frac{4}{\sqrt{3}}$ and I can work backwards geometrically from this answer given in the book but I want to know if you can work forward from the information given only by the initial problem using geometry.

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    $\begingroup$ Hints: (1) What is the longest line segment you can fit into a cube? (2) What is the longest line segment you can fit in a sphere? $\endgroup$
    – celtschk
    Commented Jul 28, 2016 at 15:30
  • $\begingroup$ BTW, "the answer consists of all like edges" — a cube by definition has all like edges. $\endgroup$
    – celtschk
    Commented Jul 28, 2016 at 15:36
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    $\begingroup$ Starting from first principles, there is no guarantee that the maximum box is a cube. That has to be proven. $\endgroup$
    – marty cohen
    Commented Jul 28, 2016 at 16:23
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    $\begingroup$ @martycohen: The title says "maximum volume of cube". I didn't notice that this changed to "box" in the actual question. $\endgroup$
    – celtschk
    Commented Jul 29, 2016 at 16:23

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Transpose the problem to two dimensions to better see what's going on. What is the largest square we can fit inside a circle of radius two?

Question: What is the longest line inside a circle of radius two?

Notice that the largest rectangle that can be put inside a circle is a square whose diagonal is a diameter of the circle. Now we move this reasoning up to three dimensions.

In three dimensions, we know we want some sort of cube of side length $x$ with volume $x^3$. Again, the diagonal of the cube must coincide with the longest distance in the sphere, namely the diameter of $4$. Now apply the pythagorean theorem twice to the cube to find the diagonal of the cube: $$x^2 + x^2 + x^2 = (4)^2 \implies x = \frac{4}{\sqrt{3}} \implies x^3 = \frac{64}{3\sqrt{3}}$$ So we are done. The key thing to remember from this is that working one dimension lower often gives you a clear path to follow in the higher dimension. In this case, we see that we can take a cross section of the sphere in question and just work with a circle, as the logic transfers directly.

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  • $\begingroup$ To specify my question more, I was seeking more of a geometric approach relationship besides a geometric explanation why a Lagrange algebraic equation works. (I knew about the square and cube relationship). Indeed the Lagrange shows that X, Y, Z are all the same and I know geometrically why they are so but my question is how can we solve using triangles and geometry? You stated to apply the Pythagorean theorem twice to the cube to find the edge X of the cube but Its hard for me to visualize the two triangles involved and how their angles were derived besides using the Lagrange/Algebraic set. $\endgroup$
    – Visualize It not Algerbrize It
    Commented Jul 28, 2016 at 18:05
  • $\begingroup$ @VisualizeItnotAlgerbrizeIt Does this image clarify it? The two triangles are one across the face of the cube, one through the center. $\endgroup$
    – Legendre
    Commented Jul 28, 2016 at 18:08
  • $\begingroup$ Im sorry for the previous comment if you read it, I deleted it. Just saw your image. Thank you, Perfectly Explained. Thanks!! Good job! $\endgroup$
    – Visualize It not Algerbrize It
    Commented Jul 28, 2016 at 19:04

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