Vtyjkyuk

I've been trying to solve this problem with a couple of methods so far (including Lagrange multipliers) but for each method I end up with an unsolvable equation.

My first approach (Lagrange multipliers) starts off with optimizing $x^2+y^2+z^2$ constrained to $left(left|dfracxaright|^r+left|dfracybright|^rright)^fractr+left|dfraczcright|^t-1 = 0$ where $a$, $b$, $c$, $r$, and $t$ are variables that tweak the shape of the superellipsoid. From there, I go from this:

$$
beginbmatrix
dfracpartialpartial xleft(x^2 + y^2 + z^2 - lambdaleft(left(left|dfracxaright|^r+left|dfracybright|^rright)^fractr+left|dfraczcright|^t-1right)right)\
dfracpartialpartial yleft(x^2 + y^2 + z^2 - lambdaleft(left(left|dfracxaright|^r+left|dfracybright|^rright)^fractr+left|dfraczcright|^t-1right)right)\
dfracpartialpartial zleft(x^2 + y^2 + z^2 - lambdaleft(left(left|dfracxaright|^r+left|dfracybright|^rright)^fractr+left|dfraczcright|^t-1right)right)\
dfracpartialpartiallambdaleft(x^2 + y^2 + z^2 - lambdaleft(left(left|dfracxaright|^r+left|dfracybright|^rright)^fractr+left|dfraczcright|^t-1right)right)
endbmatrix
=
beginbmatrix
0\0\0\0
endbmatrix
$$
to
$$
beginbmatrix
2x - dfractrlambdaleft(left(dfracxaright)^r+left(dfracybright)^rright)^dfract-rrdfracx^r-1a^r\
2y - dfractrlambdaleft(left(dfracxaright)^r+left(dfracybright)^rright)^dfract-rrdfracy^r-1b^r\
2z-dfractlambdac^tz^t-1\
left(left(dfracxaright)^r+left(dfracybright)^rright)^fractr+left(dfraczcright)^t-1
endbmatrix
=
beginbmatrix
0\0\0\0
endbmatrix
$$
to
$$
begincases
2x - dfractr2z^2-tc^tt^-1left(left(dfracxaright)^r+left(dfracybright)^rright)^dfract-rrdfracx^r-1a^r=0\
2y - dfractr2z^2-tc^tt^-1left(left(dfracxaright)^r+left(dfracybright)^rright)^dfract-rrdfracy^r-1b^r=0\
left(left(dfracxaright)^r+left(dfracybright)^rright)^fractr+left(dfraczcright)^t=1
endcases
$$
where I'm pretty much stuck, and sympy won't help.

My second attempt is converting the implicit equation of a superellipsoid into its polar form, then using implicit differentiation to get $dfracpartial rpartialtheta = 0$ and $dfracpartial rpartialtheta=0$. I get:
$$
left(left(dfracr sintheta cosphiaright)^n + left(dfracr sinphi sinthetabright)^nright)^fractn + left(dfracr costhetacright)^t = 1
$$
and sympy gives me:
$$
begincases
fracr left(left(fracr cosleft (theta right )cright)^t sin^2left (theta right ) - left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractn cos^2left (theta right )right)left(left(fracr cosleft (theta right )cright)^t + left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractnright) sinleft (theta right ) cosleft (theta right ) = 0 \
fracr left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n sin^2left (phi right ) - left(fracr sinleft (phi right ) sinleft (theta right )bright)^n cos^2left (phi right )right) left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractnleft(left(fracr cosleft (theta right )cright)^t left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr cosleft (theta right )cright)^t left(fracr sinleft (phi right ) sinleft (theta right )bright)^n + left(fracr sinleft (theta right ) cosleft (phi right )aright)^n left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractn + left(fracr sinleft (phi right ) sinleft (theta right )bright)^n left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractnright) sinleft (phi right ) cosleft (phi right ) = 0
endcases
$$
which boils down to:
$$
begincases
left(fracr cosleft (theta right )cright)^t sin^2left (theta right ) = left(left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^nright)^fractn cos^2left (theta right ) \
begincases
left(fracr sinleft (theta right ) cosleft (phi right )aright)^n sin^2left (phi right ) = left(fracr sinleft (phi right ) sinleft (theta right )bright)^n cos^2left (phi right ) \
or \
left(fracr sinleft (theta right ) cosleft (phi right )aright)^n + left(fracr sinleft (phi right ) sinleft (theta right )bright)^n = 0
endcases
endcases
$$
at which point sympy won't help once again, and I've no idea how to solve for $theta$ and $phi$.
Anyways, has someone got an alternative method for solving this? Or is it just me who can't solve the equations?

Use the change of variable

$$left(frac xaright)^r=rho^r/tcos^2theta,
\left(frac ybright)^r=rho^r/tsin^2theta.$$

Then the constraint becomes

$$rho+left(frac zcright)^t-1=0$$ or

$$z=c(1-rho)^1/t.$$

The function to be maximized is now

$$rho^2/t(a^2cos^4/rtheta+b^2sin^4/rtheta)+c^2(1-rho)^2/t.$$

Differentiating on $theta$ and canceling,

$$-a^2sinthetacos^4/r-1theta+b^2costhetasin^4/r-1theta=0$$ or

$$tan^4/r-2theta=fraca^2b^2$$ gives you the solutions in $theta$.

And differentiating on $rho$,

$$rho^2/t-1(a^2cos^4/rtheta+b^2sin^4/rtheta)-c^2(1-rho)^2/t-1$$

gives

$$frac1-rhorho=left(fraca^2cos^4/rtheta+b^2sin^4/rthetac^2right)^t/(2-t).$$