# Maximize Volume of a Rectangular Box

by dtl42
Tags: maximize, rectangular, volume
 P: 117 1. The problem statement, all variables and given/known data Find the dimensions of the rectangular box of largest volume that can be inscribed in a sphere of radius 1. 2. Relevant equations v=w*l*h, Set the partials equal to 0, then solve a system, etc. 3. 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,706 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.
 Mentor P: 20,431 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.

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