Question

show that the maximum value of a rectangular parallelopiped enclosed in the ellipsoid x^2/a^2+y^2/b^2+z^2/c^2=1 is 8abc/3 root 3​

Answer 1

Answer:

Let

h(x)=maximum value=8xyz+f(x2/a2+y2/b2+z2/C2+1)

where f =langragian multiplier.

Then differintiate partialy with respect to x

8yz+2fx/a2=0

1/a2=(−8yz/2fx)

similarily

1/b2=(−8zx/2fy)

1/c2=(−8xy/2fz)

x2/a2+y2/b2+z2/c2=1

⇒f=−12xyz

put f value in eqn(2)

x=a/

√

3

similarily

y=b/

√

3

z=c/

√

3

⇒ the largest volume of parallelopiped inscribed in ellipsoid

=8xyz=8(a/

√

3

)(b/

√

3

)(c/

√

3