I have a problem of determining the maximum volume of a cuboid inscribed in an elipsoid.
I would like to solve it without Lagrange multipliers.
I have the function $f = 8xyz$ representing the volume of the cuboid and the elipsoid constraint efoution $g = \dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c}=1$
I've tried $df\wedge dg=0$, but I get an unsolvable set of equations :
z=c/b y
z=c/a x
y=b/a x
Can this even be solved with this differential form approach?
I just took the differentials of f and g so that
df = 8dx yz + 8 dy xz + 8dz xy
and similarly for dg then just took the outer product.
Or do I have to find the multipliers explicitly?
I'm going to expand on my comments because I hope this will be helpful not only to you, but also to others. I won't work out all the algebra to actually solve the problem.
You have \begin{align*} df &= 8(yz\,dx + xz\,dy + xy\,dz) \quad\text{and}\\ dg &= 2\left(\frac xa dx + \frac yb dy + \frac zc dz\right). \end{align*}
Thus, using skew-commutativity of wedge product, $$\frac1{16} df\wedge dg = \big(\frac{y^2z}b-\frac{x^2z}a\big)dx\wedge dy + \big(\frac{yz^2}c-\frac{x^2y}a\big)dx\wedge dz + \big(\frac{xz^2}c-\frac{xy^2}b\big)dy\wedge dz.$$ Since $dx\wedge dy$, $dx\wedge dz$, and $dy\wedge dz$ are linearly independent, in order to have $df\wedge dg=0$, all three coefficients must vanish, and we get the system of equations \begin{align*} z(ay^2-bx^2) &= 0,\\ y(az^2-cx^2) &= 0,\\ x(bz^2-cy^2) &= 0. \end{align*}
Where did you get $df\wedge dg = 0$? Well, this equation expresses the fact that $df$ and $dg$ become linearly dependent at the desired points. This is equivalent to the classical statement that $\nabla f$ and $\nabla g$ become linearly dependent, hence scalar multiples. Since the overriding assumption is that $\nabla g\ne 0$ on the constraint set, this is equivalent to the condition that $\nabla f=\lambda\nabla g$ for some scalar $\lambda$. How does the usual solution go?
We have $$\nabla f = 8(yz,xz,xy) \quad\text{and}\quad \nabla g = 2\big(\frac xa,\frac yb,\frac zc\big).$$ Saying that $\nabla f =\lambda \nabla g$ says (I'm ignoring the numerical factors by replacing $\lambda$ with $4\lambda$) $$yz = \lambda\frac xa, \quad xz = \lambda\frac yb, \quad xy = \lambda \frac zc.$$ Now we eliminate $\lambda$ by taking ratios. Assume for the moment none of $x,y,z$ is $0$. $$(\lambda = )\frac{yz}{\frac xa} = \frac{xz}{\frac yb} = \frac{xy}{\frac zc},$$ or $$\frac{ayz}x = \frac{bxz}y = \frac{cxy}z.\tag{$\star$}$$ By cross-multiplying we obtain $$ay^2z=bx^2z, \quad ayz^2=cx^2y, \quad bxz^2=cxy^2.$$ Look familiar? (By cross-multiplying, in this case, we get the equations that allow the denominators to be $0$, and those are indeed cases we would check separately if we stopped at ($\star$) to work with the equations. For example, the first gives $z=0$ or $a\frac yx = b\frac xy$.)
In summary, the two approaches are identical. In notation that is used various places in the world, we replace $\nabla f\times\nabla g = 0$ (which, in $\Bbb R^3$, is one way of saying the two vectors are linearly dependent) with $\nabla f\wedge\nabla g = 0$, and this corresponds to the dual statement $df\wedge dg=0$.