Question

Find the volume of the largest rectangular solid which can be inscribed in the

ellipsoid

x

2

a2

+

y

2

b

2

+

z

2

c

2

= 1

Answer 1

Solution:

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

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}=1$

$\therefore$  The lengths of three co-terminous edges of the rectangular solid are 2 x, 2 y, 2 z .

$\therefore$  Volume $\quad \mathrm{V}=2 x .2 y .2 z=8 x y z$

We are make V maximum

Let $\quad \mathbf{V}=f(x, y, z)=8 x y z$

Let $\phi(x, y, z)=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}}-1=0$

Art 5 ;

Now using equations ;  

$\frac{\partial f}{\partial x}+\lambda \frac{\partial \phi}{\partial x}=0\ gives\ 8 y z+\lambda \frac{2 x}{a^{2}}=0 \ldots(2) \\ \frac{\partial f}{\partial y}+\lambda \frac{\partial \phi}{\partial y}=0\ gives\ 8 x z+\lambda \frac{2 y}{b^{2}}=0 \ldots(3)$

and $\frac{\partial f}{\partial z}+\lambda \frac{\partial \phi}{\partial z}=0\ gives\ 8 x y+\lambda \frac{2 z}{c^{2}}=0 ...(4)$

Multiplying equations (2), (3) and (4) by x, y and z respectively and adding them, we have

$\begin{array}{rlr} & 8 y z-12 x y z \cdot \frac{2 x}{a^{2}}=0 & \\ & \text { or } 8 a^{2} y z-24 x^{2} y z=0 & \text { or } & 8 y z\left(a^{2}-3 x^{2}\right) & =0 \\ \text { But } \quad 8 y z \neq 0 . & & \therefore & a^{2}-3 x^{2} & =0 \\ \text { i.e., } \quad 3 x^{2}=a^{2} & \text { or } & x^{2} & =\frac{a^{2}}{3} \\ \therefore \quad & x=\frac{a}{\sqrt{3}} & & \end{array}$

Similarly, by putting the values of  $\lambda$  in Eqns. (3) and (4), we get

$y=\frac{b}{\sqrt{3}} \quad \text { and } \quad z=\frac{c}{\sqrt{3}} \\\therefore \quad \text { Maximum volume }=8 x y z=8 \cdot \frac{a}{\sqrt{3}} \cdot \frac{b}{\sqrt{3}} \cdot \frac{c}{\sqrt{3}}=\frac{8 a b c}{3 \sqrt{3}} .$