Let (x, y, z) be the co-ordinates of a vertex of the rectangular parallelopiped in the positive octant, which is inscribed in the ellipsoid
a2x2+b2y2+c2z2=1
∴ The lengths of three co-terminous edges of the rectangular solid are 2 x, 2 y, 2 z .
∴ Volume V=2x.2y.2z=8xyz
We are make V maximum
Let V=f(x,y,z)=8xyz
Let ϕ(x,y,z)=a2x2+b2y2+c2z2−1=0
Art 5 ;
Now using equations ;
∂x∂f+λ∂x∂ϕ=0 gives 8yz+λa22x=0…(2)∂y∂f+λ∂y∂ϕ=0 gives 8xz+λb22y=0…(3)
and ∂z∂f+λ∂z∂ϕ=0 gives 8xy+λc22z=0...(4)
Multiplying equations (2), (3) and (4) by x, y and z respectively and adding them, we have
But 8yz=0. i.e., 3x2=a2∴8yz−12xyz⋅a22x=0 or 8a2yz−24x2yz=0 or x=3a or ∴x28yz(a2−3x2)a2−3x2=3a2=0=0
Similarly, by putting the values of λ in Eqns. (3) and (4), we get
y=3b and z=3c∴ Maximum volume =8xyz=8⋅3a⋅3b⋅3c=338abc.
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