# Math Help - volume corrections

1. ## volume corrections

largest box with edges that can be inscribed into the ellipsoid $x^2/64+y^2/64+z^2/36=1$
I again used the larange multiplier using v=8xyz and my g function.

2. Dear JaneP.,

I do not like the Langrange multiplication method, so I suggest you to stay at the elementary way: from constraint(now the equation of the ellipsoid) we express one of the variations and substitute into V function.So we get a two-variant function, we can find the maximum with the usage of derivation.

Trick: use V^2 insteed V so you don't get square root.

3. Hello, JaneP!

Largest box with edges that can be inscribed into the ellipsoid $\frac{x^2}{64}+\frac{y^2}{64}+\frac{z^2}{36} \:=\:1$

I again used the Larange multiplier using $V\,=\,8xyz$ and my $g$ function.
. . Due to the symmetry, we can work in the first octant, so that: ${\color{blue}x,y,z \geq 0}$

We have: . $f(x,y,z,\lambda) \;=\;8xyz + \lambda\left(\frac{x^2}{64} + \frac{y^2}{64} + \frac{z^2}{32} - 1\right)$

Equate the partial derivatives to zero . . .

. . $\begin{array}{cccccc}f_x &=& 8yz + \frac{\lambda x}{32} &=& 0 & {\color{blue}[1]}\\ \\[-4mm]
f_y &=& 8xz + \frac{\lambda y}{32} &=& 0 & {\color{blue}[2]}\\ \\[-4mm]
f_z &=& 8xy + \frac{\lambda z}{16} &=& 0 & {\color{blue}[3]}\\ \\[-4mm]
f_{\lambda} &=& \frac{x^2}{64} + \frac{y^2}{64} + \frac{z^2}{32} - 1 &=& 0 & {\color{blue}[4]} \end{array}$

From [1], we have: . $\lambda \:=\:-\frac{256yz}{x}\;\;{\color{blue}[4]}$

From [2], we have: . $\lambda \:=\:-\frac{256xz}{y}\;\;{\color{blue}[5]}$

From [3], we have: . $\lambda \:=\:-\frac{128xy}{z}\;\;{\color{blue}[6]}$

Equate [4] and [5]: . $-\frac{256yz}{x} \:=\:-\frac{256xz}{y} \quad\Rightarrow\quad y^2\:=\:x^2\;\;{\color{blue}[7]}$

Equate [4] and [6]: . $-\frac{256yz}{x} \:=\:-\frac{128xy}{z} \quad\Rightarrow\quad z^2 \:=\:\tfrac{1}{2}x^2\;\;{\color{blue}[8]}$

Substitute [7] and [8] into [4]: . $\frac{x^2}{64} + \frac{x^2}{64} + \frac{\frac{1}{2}x^2}{32} -1 \:=\:0$

. . $\frac{3}{64}x^2 \:=\:1 \quad\Rightarrow\quad x^2 \:=\:\frac{64}{3} \quad\Rightarrow\quad y^2 \:=\:\frac{64}{3} \quad\Rightarrow\quad z^2 \:=\:\frac{32}{3}$

Therefore: . $\begin{Bmatrix}x &=& \sqrt{\dfrac{64}{3}} &=& \dfrac{8\sqrt{3}}{3} \\
y & =& \sqrt{\dfrac{64}{3}} &=& \dfrac{8\sqrt{3}}{3} \\
z &=&\sqrt{\dfrac{32}{3}} &=& \dfrac{4\sqrt{6}}{3} \end{Bmatrix}$