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Math Help - volume corrections

  1. #1
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    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.
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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 \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]<br />
f_y &=& 8xz + \frac{\lambda y}{32} &=& 0 & {\color{blue}[2]}\\ \\[-4mm]<br />
f_z &=& 8xy + \frac{\lambda z}{16} &=& 0 & {\color{blue}[3]}\\ \\[-4mm]<br />
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} \\ <br />
y & =& \sqrt{\dfrac{64}{3}} &=& \dfrac{8\sqrt{3}}{3} \\<br />
z &=&\sqrt{\dfrac{32}{3}} &=& \dfrac{4\sqrt{6}}{3} \end{Bmatrix}

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