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I have to solve the following problem.

Task-description: A sphere with the radius of $10$ contains a pyramid with the maximal lateral surface ($M_{max}$). Find $M_{max}$.

enter image description here

I need two terms. One of them is the following, but I don't find a second one. enter image description here

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I am not too sure what the problem asks. If you take smaller and smaller triangular faces, you get a pyramid that is flatter and flatter, tending to a square in fact, a 2-dimensional set. Then the maximal square is contained in the same plane as the boundary of the scoop, and therefore has sides of length $10\sqrt 2$ and area $200$. – plm May 14 '12 at 9:02
The topic is extremal-combinatoric. I need the maximal lateral surface from the pyramid which is in the ball. For that, we need two terms. One of that is the main-term which is in my second picture. – alexander-fire May 14 '12 at 9:15
@user694501: I think you need to explain more about your solution in the second picture. – Gigili May 14 '12 at 9:17
Hmm, I guess the problem asks for the maximum total area of the pyramid. And anyway you are certainly allowed to put the pyramid upside down. So you get the square face of area 200 plus 4 triangular faces of base the same length as the square and height $5\sqrt{6}=\sqrt{(5\sqrt 2)^2+10^2}$. The triangular faces have area $10\sqrt 2\cdot 5\sqrt{6}=100\sqrt 3$, so you have a total area of $200+400\sqrt 3\approx 900$. – plm May 14 '12 at 9:23
To prove that the symmetric solution I proposed is maximal should not be too hard. And for Fourier analysis fans there must be explanations in terms of that and perhaps interpretations in terms of volumes on Lie groups with their Haar measure. I add this just to say something grandiose and surely wrong, my trademark -at least there is the Lebesgue measure involved :). – plm May 14 '12 at 9:30

Let $S=(0,0,1)$, let $P$ be one of the base points, and let $\alpha:=\angle(OSP)$. Then the length of the ascending edges is $2\cos\alpha$, and the length of the base diagonal is $2\sin(2\alpha)$. It follows that the side walls of the pyramid are isosceles triangles with base $\sqrt{2}\sin(2\alpha)$ and height $$h=\sqrt{4\cos^2\alpha-\sin^2(2\alpha)/2}\ .$$ Therefore the total lateral surface $M$ is given by $$M=4{1\over2}\sqrt{2}\sin(2\alpha)\, h=8\sin\alpha\ \cos^2\alpha\sqrt{2-\sin^2\alpha}\ .$$ Putting $\sin\alpha=:s$ we now have to maximize the function $$f(s):=s(1-s^2)\sqrt{2-s^2}\qquad(0\leq s\leq 1)\ .$$ Doing the computations gives $s_{\rm opt}=\sqrt{1-1/\sqrt{2}}$.

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Let $O$ be the midpoint of the line segment $AB$. (I use the same notation for the points as in your picture.)

You have correctly noticed that $$h_a^2=h^2+\frac{a^2}4,$$ which is equation obtained from the right triangle $SNO$.

You can use equation $$r^2=(h-r)^2+\frac{a^2}2$$ which you obtain from the right triangle $MNA$. (Just notice that the distance $|MN|$ is $h-r$.)

The above equation is equivalent to $$\begin{align} r^2&=h^2-2hr+r^2+\frac{a^2}2\\ 0&=h^2-2hr+\frac{a^2}2\\ 2hr&=h^2+\frac{a^2}2\\ \frac{a^2}2&=2hr-h^2=h(2r-h) \end{align}$$

So you have $h_a^2=h^2+\frac{a^2}4=h^2+\left(hr-\frac{h^2}2\right)=hr+\frac{h^2}2$.

Maximizing $2ah_a$ is equivalent to maximizing $$(ah_a)^2=(2hr-h^2)(2hr+h^2).$$ Or, if you denote $t=\frac{h}r$, you get $$(ah_a)^2=r^4(2t-t^2)(2t+t^2),$$ which reduces the problem to maximizing $(2t-t^2)(2t+t^2)$.

BTW are you sure you want to maximize $2ah_a$ and not $2ah_a+a^2$? (The second one is the surface including the base. It seems that this would be a more difficult problem.)

Note that you can get $\frac{a^2}2=2hr-h^2=h(2r-h)$ also from triangle $SAS'$, where $S'$ is the point opposite to $S$; if you use right triangle altitude theorem. (The triangle $SAS'$ is right triangle according to Thales's theorem. The hypotenuse $|SS'|=2r$ is divided into two parts of lengths $h$ and $2r-h$.)

Added later:

The function $f(t)=(2t-t^2 )(2t+t^2)=t^2(4-t^2)$ attains maximal value $f(t)=4$ for $t^2=2$, i.e. $t=\sqrt2$. (See Wolframalpha. This can be also seen directly, by maximazing the function $t^2(4-t^2)$ -- perhaps with the substitution $s=t^2$ -- or by by applying AM-GM inequality as $t^2(4-t^2)\le \frac{t^2+(4-t^2)}2=2$.)

This corresponds to $(ah_a)^2=4r^2$, $ah_a=2r$ and the lateral surface area $$M=2ah_a=4r.$$ We also have $h=tr=\sqrt2r$.

If we compare this with Christian Blatter's solution, he gets maximal value $f(s_{\rm opt})=\frac12$, see wolframalpha. This gives $M=8f(s_{\rm opt})=4$.

We can also notice that the optimal angle is $\frac\pi8$, since $s_{\rm opt}=\sqrt{1-1/\sqrt{2}}=\sqrt{\frac{2-\sqrt2}{\sqrt2}}=\sin\frac\pi8$, see Wikipedia.

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