Find Maximum Volume of a rectangular box that is inscribed

Carol Gates 2021-09-19 Answered
Find Maximum Volume of a rectangular box that is inscribed in a sphere of radius r.
The part i do not get the solutions manual got 2x, 2y, and 2z as the volume. Can you explain how the volume of the cube for this problem is v=8xyz?
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Expert Answer

Bertha Stark
Answered 2021-09-20 Author has 96 answers
Step 1
Recall that the volume of a box is V=l×w×h. The volume of the box inscribed inside a sphere where (x, y, z) is the point in octant 1 where the box touches, the sphere will be V=(2x)(2y)(2z)=8xyz. Since this point lies on a sphere, it must satisfy x2+y2+z2=r2. We have two equations in which we can get rid of one variable.
V=8xyz
z=V8xy
x2+y2+z2=r2
z=(r2x2y2) (no ± because in octant 1, z is positive)
V8xy=(r2x2y2)
V=8xy(r2x2y2)
This is just an optimization problem now.
fx=8(y(r2x2y2)(x2)y(r2x2y2))=0
fy=8(y(r2x2y2)(y2)x(r2x2y2))=0
The solution to these sets of equations are (r3, r3) (there's also (0, 0) but that's obvious not a maximum point)
The maximum volume therefore is:
V=8xyz
=8(r3)(r3)(r3)
=8r3(33)
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