Answer to Question #104409 in Calculus for Effiong israel

Question

Using Lagrange Obtain the dimensions of a box of maximum volume that can be inscribed in a ellipsoid x²/a²+y²/b²+z²/c²=1

Answer 1

$x²/a²+y²/b²+z²/c²=1$ ----------(1)

This is an ellipsoid that is symmetrical about the origin.

Thus, the box of maximum volume will also be symmetrical about the origin and it's vertices will lie on the ellipsoid's surface.

Let one of the vertices be (x,y,z), where x,y, z>0.

Thus, the box has the dimensions 2x, 2y and 2z.

$\implies V_b=(2x)(2y)(2z)=8xyz=max$ .

From (1), $z=c(1-x²/a²-y²/b²)^{1/2}$

$\implies V_b=8cxy(1-x²/a²-y²/b²)^{1/2}$

$\delta V_b/ \delta x=8cy(1-x²/a²-y²/b²)^{1/2}+8cxy(-x/a^2)(1-x²/a²-y²/b²)^{-1/2}$

$=(1-x²/a²-y²/b²-x²/a²)8cy(1-x²/a²-y²/b²)^{-1/2}$

$=(1-2x²/a²-y²/b²)8cy(1-x²/a²-y²/b²)^{-1/2}$

$\delta V_b/ \delta y=8cx(1-x²/a²-y²/b²)^{1/2}+8cxy(-y/b^2)(1-x²/a²-y²/b²)^{-1/2}$

$=(1-x²/a²-y²/b²-y²/b²)8cx(1-x²/a²-y²/b²)^{-1/2}$

$=(1-x²/a²-2y²/b²)8cx(1-x²/a²-y²/b²)^{-1/2}$

Now, for maximum volume, $\delta V_b/ \delta x=0; \delta V_b/ \delta y=0$ ;thus we get;

$1-2x²/a²-y²/b² \implies 2x²/a²+y²/b²=1$

$1-x²/a²-2y²/b² \implies x²/a²+2y²/b²=1$

Solving these two equations we get;

$3x^2/a^2=1 \implies x=a/ \sqrt[]{3}$

$\implies y=b/ \sqrt{3}$

$\implies z=$ $c(1-x²/a²-y²/b²)^{1/2}=c/ \sqrt {3}$

$\implies V_b(max)=8abc/3 \sqrt {3}$ .