prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube?

Question

please explain this question by a diagram

Answer 1

Let the sphere be x^2+y^2+z^2=1, and the rectangular solid's vertices be (±x,±y,±z). For a cube we have x=y=z=1/sqrt(3).

The volume of the solid is 8xyz. Now z=sqrt(1-x^2-y^2) so you want to maximize

A(x,y) = 8xy*sqrt(1-x^2-y^2)

I hate differentiating this kind of stuff so I'll square it. (Finding the x and y that maximizes the square of a function is the equivalent of maximizing the function itself, because the square is an increasing function over positive values.) So.

A^2 = 64x^2y^2(1-x^2-y^2)
= 64(x^2y^2 - x^4y^2 - x^2y^4)

Differentiating with respect to x and y,
d(A^2)/dx = 64(2xy^2 - 4x^3y^2 - 2xy^4)
d(A^2)/dy = 64(2x^2y - 2x^4y - 4x^2y^3).

Rather than solve this monster, I will just plug the coordinates of the cube, x=y=1/sqrt(3) and see what happens...

d(A^2)/dx = 64(2/sqrt(3)^3 - 4/sqrt(3)^5 - 2/sqrt(3)^5)
= 64/sqrt(3) * (2/3 - 4/9 - 2/9)
= 0.
Similarly, d(A^2)/dy = 0.

So the point x=y=1/sqrt(3) is a maximum for the function A(x,y)=8xyz. This results in z=1/sqrt(3) as well - the cube. The resulting volume is 8/(3*sqrt(3)).

Formally, you also need to show that the second derivatives are negative to prove that the point x=y=1/sqrt(3) is a local maximum and not something like a saddle point but I've already written enough...

Answer 2

This is just the three dimentional version of the problem of finding the maximum area of a rectangle that is drawn in a circle.

A simple version of the logic would be that as one side approaches zero the area approaches zero. Since that is true for both sides the maximum area must be half way between, which is where they are the same length or a square.

You could use similar logic with the three sides of a rectangular solid in a sphere.

It is impossible to draw a diagram on Yahoo answers.

Answer 3

um your telling us to perform a higher level proof, how can we prove with a diagram we can't draw one here. and there are so many theorms involved and stuff. someone that's really determined about mathematics will do this for you. i think your intelligent enough to do it yourselve. here just to give you a clue, prove it by using a indirect proof another clue is that a cube when inscribed in a sphere touches it with it's endpoints so the longest diagnal would give you the diameter of the sphere.

Answer 4

i can't depict the model using a digram

i can describe it

when a rectangle's stretched u 'd observe that it resemble a wire
i.e., u observe that surface area is icreased
n the vol decreases

if it's modified into a cube the surface area decreases n the vol increases

a sphere has the least surface rea n highest vol for the given dimensions

try this using an eraser n a rubber ball


u'll get the required answer

Answer 5

a cyclic rectangle with maximum volume is always a square
there fore a rectangular parallelopiped with maximum volume is always a cube
diagram in yahoo anwers is not possible

Answer 6

Do you propose optimal quantity of an oblong field it somewhat is inscribed around a sphere of radius r? if so, the field could be a sq. or the question is unsolvable. no count if it somewhat is than the quantity is (2r)^2.

Answer 7

Diagrams not possible.