Question

Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

Show that the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.

Let's assume that a sphere of radius 'r' is inscribed with a rectangular solid of maximum volume.

Let's call the length, width, and height of the rectangular solid as 'l', 'w', and 'h' respectively.

Since the rectangular solid is inscribed inside the sphere, each vertex of the rectangular solid must be at a distance of 'r' from the center of the sphere.

Therefore, we can write the following three equations:

$(l/2)^2 + (w/2)^2 + (h/2)^2 = r^2$

$(l/2)^2 + (w/2)^2 = (r - h/2)^2$

$(l/2)^2 + (h/2)^2 = (r - w/2)^2$

Squaring the third equation and adding it to twice the square of the second equation, we get:

$2(l/2)^2 + 2(w/2)^2 + 2(h/2)^2 = 3r^2$

Comparing this with the first equation, we get:

$2r^2 = 3r^2,$

which is a contradiction. Therefore, the only possibility is that all three dimensions, l, w, and h, are equal

. Thus, the rectangular solid of maximum volume that can be inscribed in a sphere is a cube.

To know more about the concept please go through the links

brainly.in/question/1116370

brainly.in/question/6921511

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