Maximize Volume of a Rectangular Box

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To maximize the volume of a rectangular box inscribed in a sphere of radius 1, the box's dimensions must be determined using the constraint of the sphere's equation, x^2 + y^2 + z^2 = 1. Setting up a coordinate system with the sphere's center at (0,0,0) simplifies the problem, allowing for the analysis of the box in the first octant. The volume of the box can be expressed as V = xyz, and by substituting z from the sphere's equation, the volume function can be reduced to two variables. Taking partial derivatives with respect to x and y and setting them to zero will help identify the maximum volume. The approach emphasizes the importance of proper coordinate system setup and understanding geometric constraints in optimization problems.
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Homework Statement


Find the dimensions of the rectangular box of largest volume that can be inscribed in a sphere of radius 1.


Homework Equations


v=w*l*h, Set the partials equal to 0, then solve a system, etc.


The Attempt at a Solution


I'm really just unsure of the constraints that might arise when inscribing a box in a sphere, I'm fairly confident about the rest of the process.
 
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One thing students seem to have trouble realizing is that applications don't typically come with coordinate systems attached! You have a sphere, sittin there in space, with a rectangle inscribed in it. You can't write equations until you have set up a coordinate system. The obvious thing, I think, is to choose your coordinate system so that (0,0,0) is at the center of the sphere and then the equation of the sphere is x^2+ y^2+ z^2= 1.

That still leaves the orientation of the axes- again, it would strike me as simplest to choose the axes parallel to the edges of the box. Now, one corner of the box will be in the first octant, (x, y, z) with x, y, and z positive, and, of course, x^2+ y^2+ z^2= 1. It should be easy, using the fact that the edges of the are parallel to the axes, and using the symmetry of the sphere, to write down the coordinates of the other 7 corners and so find the lengths of the edges and the volume as a function of x, y, and z.
 
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For the sphere of radius 1, with center at (0, 0, 0), the equation is x^2 + y^2 + z^2 = 1.

To simplify things, you can work with just the portion of the sphere in the first octant (i.e., x, y, z >= 0) and the one-eighth of the rectangular box that is in this octant.

For this box, the vertex opposite the one at the origin is at (x0, y0, z0) on the sphere. No other corners of the box touch the sphere.

You want to find the max. value of V = xyz, subject to the constraint that x^2 + y^2 + z^2 = 1. From the latter equation you can solve for z to make your volume a function of x and y alone. Then you can take partials wrt x and y and use them to find the max. volume.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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