Question

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)

Answer 1

let (x,y,z) be the coordinates of the corner of the box that is inscribed in an ellipsoid. Thus $x,y,z \geq0$ and equation of ellipsoid is

$\frac{x^2}{9}+\frac{y^2}{16}+\frac{z^2}{36}=1$

$\frac{z^2}{36}=1-\frac{x^2}{9}-\frac{y^2}{16}$

so $z=6\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}$

The volume of the box is represented as

$V=(2x)(2y)(2z)=48xy\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}$

For $x\geq0,y\geq0, and \ (\frac{x^2}{a^2})+(\frac{y^2}{b^2})\leq1$

For analysis it is easier to deal with $V^2\ than\ V$ as it becomes more complex to find the partial derivative in terms of x and y

$V^2=2304[x^2y^2-\frac{x^4y^2}{9}-\frac{x^2y^4}{16})$

Since $V=0 \ if \ x=0 \ or \ y=0 \ or\ (\frac{x^2}{a^2})+(\frac{y^2}{b^2})=1,$

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

$\frac{\delta V^2}{\delta x}=2304[2xy^2-\frac{4x^3y^2}{9}-\frac{2x^2y^4}{16})$

$=4608xy^2(1-\frac{2x^2}{9}-\frac{y^2}{16})........(1)$

similarly

$\frac{\delta V^2}{\delta y}=4608xy^2(1-\frac{x^2}{9}-\frac{2y^2}{16})........(2)$

Now we will equate the above partial derivatives to zero so as to find critical points

$(1-\frac{2x^2}{9}-\frac{y^2}{16})=0\\\frac{2x^2}{9}+\frac{y^2}{16}=1........(3)$

$(1-\frac{x^2}{9}-\frac{2y^2}{16})=0\\\frac{x^2}{9}+\frac{2y^2}{16}=1........(4)$

Now by equating (3) and (4) we get

$\frac{2x^2}{9}+\frac{y^2}{16}=1=\frac{x^2}{9}+\frac{2y^2}{16}$

which

$\frac{x^2}{9}=\frac{1}{3}\implies x=\frac{9}{\sqrt{3}}$

$\frac{y^2}{16}=\frac{1}{3}\implies y=\frac{16}{\sqrt{3}}$

The volume of the largest box that can be inscribed inside the ellipsoid is

$V=48xy\sqrt{1-\frac{x^2}{9}-\frac{y^2}{16}}$

$=48(\frac{9}{\sqrt{3}})(\frac{16}{\sqrt{3}})\sqrt{1-\frac{1}{3}-\frac{1}{3}}$

$=1330.215$