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?

Question

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?

Bertha Stark · Accepted Answer

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=(r2−x2−y2) (no ± because in octant 1, z is positive)  V8xy=(r2−x2−y2)  V=8xy(r2−x2−y2)  This is just an optimization problem now.  ∂f∂x=8(y(r2−x2−y2)−(x2)y(r2−x2y2))=0  ∂f∂y=8(y(r2−x2−y2)−(y2)x(r2−x2y2))=0  The solution to these sets of equations are (r3, r3) (there&#039;s also (0, 0) but that&#039;s obvious not a maximum point)  The maximum volume therefore is:  V=8xyz  =8(r3)(r3)(r3)  =8r3(33)

Find Maximum Volume of a rectangular box that is inscribed

Answered question

Answer & Explanation

New Questions in College Statistics