Question

Prove that a rectangular solid of maximum volume within sphere is a cube

Answer 1

Answer:

Step-by-step explanation:

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...

hope this will help you..

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