Answer to Question #274266 in Calculus for 1234

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 P=(x,y,z) be a point on the ellipsoid with x,y,z>0. Take the eight different points with

P_i​(±x,±y,±z)

These points are the vertices of a parallelepiped with the side length

2x,2y and 2z

Then, the volume parallelepiped is:

$V=2x⋅2y⋅2z=8xyz$

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

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

$V_x=48y\sqrt{1-x^2/9-y^2/16}-\frac{16x^2y}{3\sqrt{1-x^2/9-y^2/16}}=0$

$9(1-x^2/9-y^2/16)-x^2=0$

$V_y=48x\sqrt{1-x^2/9-y^2/16}-\frac{16y^2x}{3\sqrt{1-x^2/9-y^2/16}}=0$

$9(1-x^2/9-y^2/16)-y^2=0$

$x^2=y^2$

$x=y$

$9-2x^2-9x^2/16=0$

$41x^2=144$

$x=y=\sqrt{41}/12$

$V_{xx}=48y\sqrt{1-x^2/9-y^2/16}-\frac{16x^2y}{3\sqrt{1-x^2/9-y^2/16}}=$

$=-\frac{32xy}{3\sqrt{1-x^2/9-y^2/16}}$

$z=6\sqrt{1-x^2/9-x^2/16}=\sqrt{144-25x^2}/2=\sqrt{19711}/24$

$V_{max}=\frac{8\cdot41\sqrt{19711}}{144\cdot24}=13.32$