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A huge conical tank to be made from a circular piece of sheet metal of radius 10m by cutting out a sector with vertex angle theta and welding the straight edges of the straight edges of the remaining piece together. Find theta so that the resulting cone has the largest possible volume.

Specifically, the question is asked in the context of wanting derivatives, multiple max/min equations, and hopefully more calc rather than trig or geo.

I have gotten as far as using 10m as the hypotenuse for a triangle formed by the height of the cone, radius of the base of the cone, and slant. I'm not sure where to go from there, because I can't determine how to find height and/or radius, without which I'm not sure I can continue.

Let $X=(X_1,...,X_n)$ be a vector of $n$ random variables. Consider the following maximization problem:
$\underset{a,b}{max}\mathrm{C}\mathrm{o}\mathrm{v}(a\cdot X,b\cdot X)$ under the constraint that $\Vert a{\Vert }_2=\Vert b{\Vert }_2=1$.
($a\cdot X$ is the dot product between $a$ and $X$). Would it be true that there is a solution to this maximization problem such that $a=b$?
Thanks.

I am trying to solve the following question:
$\text{Maximize }f(x_1,x_2,\dots ,x_n)=2\sum _{i=1}^n{x}_i^{t}A{x}_i+\sum _{i=1}^{n-1}\sum _{j>i}^n(x_i^{t}A{x}_j+x_j^{t}A{x}_i)$
subject to
$x_i^t{x}_j=\{\begin{array}{cc}1& i=j\\ -\frac{1}{n}& i\ne j\end{array}$
where xi's are column vectors ($m\times 1$ matrices) with real entries and A is an $m\times m$ ($(n<m)$) real symmetric matrix.

From some source, I know the answer as
$f_{max}=\frac{n+1}{n}\sum _{i=1}^n{\lambda }_i^{\downarrow },$
${\lambda }_i^{\downarrow }$ being the eigenvalues of $A$ sorted in non-increasing order (counting multiplicity). But I am unable to prove it. I will appreciate any help (preferably with established matrix inequality, or Lagrange's multiplier method).

Given matrix-valued function $A:{\mathbb{R}}^d\to {\mathbb{R}}^{n\times n}$ and (tall) matrix $B\in {\mathbb{R}}^{n\times m}$, where $n>m$, let matrix-valued function $C:{\mathbb{R}}^d\to {\mathbb{R}}^{m\times m}$ be defined by
$C(x):=B^{T}A(x)B$
I would like to maximize the determinant of $C(x)$ without directly considering $C(x)$. If I maximize the determinant of $A(x)$, does that maximize the determinant of $C(x)$ as well?

If it were $m=n$, then we would have:
$det(C)=det(B{)}^{2}det(A)$
Hence, maximising $det(A)$ would maximize as a consequence $det(C)$. Does it hold something similar in the case with $n>m$? Or is it possible to define some lower bound for $det(C)$? And if so, how to prove it?

If in the previous general case it doesn't hold, maybe it could help that, in my specific case $C$ and $A$ are two positive definite matrices and $B$ is a sparse matrix of zeros and ones. $A$ is also block diagonal. The only way I came up with is using Cauchy-Binet for the determinant of the product of rectangular matrices but I remained stuck with a summation involving the principal minors of the Cholesky factorization of $A$ times minors in $B$.

Consider the example of maximizing $x^{2}yz$ under the constraint that $x^2+y^2+z^2=5$.

One way to do this is to use lagrange multipliers, solving the system of equations
$2xyz=2x\lambda$
$x^{2}z=2y\lambda$
$x^{2}y=2z\lambda$
$x^2+y^2+z^2=5$
However, couldn't you just substitute $x^2=5-y^2-z^2$ into the expression you want to maximize to get: $yz(-y^2-z^2+5)$and then just maximize that by setting the $y$ and $z$ partial derivatives equal to zero?

Then you just have to solve the arguably simpler system of equations:
$3y^{2}z+z^3=5z$
$y^3+3y{z}^2=5y$
where the $z$s and $y$s cancel out nicely on both sides.

Why is maximize by lagrange multipliers necessary when you can always substitute and maximize the resulting function?

When should you choose one over the other?

"A common theme in linear compression and feature extraction is to map a high dimensional vector $x$ to a lower dimensional vector $y=Wx$ such that the information in the vector $x$ is maximally preserved in $y$. Opten PCA is applied for this purpose. However, the optimal setting for $W$ is in generall not given by the widely used PCA. Actually, PCA is sub-optimal special case of mutual information maximisation."

Can anyone elaborate why PCA is a sub-optimal special case of mutual information maximisation ?

I need help with the following optimization problem

$max\alpha \ln (x(1-y^2))+(1-\alpha )\ln (z)$
where the maximization is with respect to $x,y,z$, subject to
$\begin{array}{rl}\alpha x+(1-\alpha )z& =C_1\\ \alpha y\sqrt{x(x+\gamma )}-\alpha x& =C_2\end{array}$
where $0\le \alpha \le 1$, $\gamma >0$, and $x,z\ge 0$, and $|y|\le 1$.

Generally, one can substitute the constraints in the objective function and maximize with respect to one parameter. The problem is that in this way things become algebraically complicated, and I believe that there is a simple solution.