0
$\begingroup$

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

  • 0
    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 $200$.2012-05-14
  • 0
    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.2012-05-14
  • 0
    @user694501: I think you need to explain more about your solution in the second picture.2012-05-14
  • 0
    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 $200+400\sqrt 3\approx 900$.2012-05-14
  • 0
    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 :).2012-05-14
  • 0
    I looked at the picture now, I thought a scoop was a half-sphere. Taking a full sphere changes the problem, but it must not really be harder.2012-05-14
  • 0
    Does your pyramid have to have a square (or quadrilateral) base? If not, a cone will have greater surface than any convex polygon. Even with a square, you should be able to write an equation connecting the side of the square and the height of the pyramid, assuming the corners of the square are on a small circle of the sphere and the peak is on the opposite pole. Then differentiate and set to zero.2012-05-24

2 Answers 2