Answer to Question #267606 in Calculus for chaitu

Question #267606

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


1
Expert's answer
2022-02-21T18:38:25-0500

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

x2a2+y2b2+z2c2=1\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 V=2x.2y.2z=8xyz\quad \mathrm{V}=2 x .2 y .2 z=8 x y z

We are make V maximum

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

Let ϕ(x,y,z)=x2a2+y2b2+z2c21=0\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 ;  

fx+λϕx=0 gives 8yz+λ2xa2=0(2)fy+λϕy=0 gives 8xz+λ2yb2=0(3)\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 fz+λϕz=0 gives 8xy+λ2zc2=0...(4)\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

8yz12xyz2xa2=0 or 8a2yz24x2yz=0 or 8yz(a23x2)=0 But 8yz0.a23x2=0 i.e., 3x2=a2 or x2=a23x=a3\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  λ\lambda  in Eqns. (3) and (4), we get

y=b3 and z=c3 Maximum volume =8xyz=8a3b3c3=8abc33.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}} .


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