What is the volume of the largest rectangular parallelepiped which can be inscribed in the ellipsoid x2/9+y2/16+z2/36=1? Show that the answer you get gives you the largest volume.(DO NOT USE LAGRANGE MULTIPLIERS)
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Expert's answer
2021-12-03T13:19:45-0500
let (x,y,z) be the coordinates of the corner of the box that is inscribed in an ellipsoid. Thus x,y,z≥0 and equation of ellipsoid is
9x2+16y2+36z2=1
36z2=1−9x2−16y2
so z=61−9x2−16y2
The volume of the box is represented as
V=(2x)(2y)(2z)=48xy1−9x2−16y2
For x≥0,y≥0,and(a2x2)+(b2y2)≤1
For analysis it is easier to deal with V2thanV as it becomes more complex to find the partial derivative in terms of x and y
V2=2304[x2y2−9x4y2−16x2y4)
Since V=0ifx=0ory=0or(a2x2)+(b2y2)=1,
the maximum value of V2 and hence of V will occur at critical point of V2 to find the critical point we we have to find the partial derivative in terms of x and y
δxδV2=2304[2xy2−94x3y2−162x2y4)
=4608xy2(1−92x2−16y2)........(1)
similarly
δyδV2=4608xy2(1−9x2−162y2)........(2)
Now we will equate the above partial derivatives to zero so as to find critical points
(1−92x2−16y2)=092x2+16y2=1........(3)
(1−9x2−162y2)=09x2+162y2=1........(4)
Now by equating (3) and (4) we get
92x2+16y2=1=9x2+162y2
which
9x2=31⟹x=39
16y2=31⟹y=316
The volume of the largest box that can be inscribed inside the ellipsoid is
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