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Thread: volume corrections

  1. #1
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    volume corrections

    largest box with edges that can be inscribed into the ellipsoid $\displaystyle x^2/64+y^2/64+z^2/36=1$
    I again used the larange multiplier using v=8xyz and my g function.
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  2. #2
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    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.
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  3. #3
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    Hello, JaneP!

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

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

    We have: .$\displaystyle 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 . . .

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


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

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

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


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

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


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

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


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

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