Find the angle between the base and the side of a pyramid.

Question

Been stuck with this problem for ages now. Please help.

The text reads: Given a pyramid with a square base and the tip of the pyramid is above where the diagonals intersect (the sides are isosceles triangles), calculate the angle between the base and the side of the pyramid so that the volume of the pyramid is the biggest. Areas of the sides is constant.

Thank you!

Answer 1

It's a constrained optimisation problem. The standard way would be to use Lagrange multipliers. But, I'll avoid to use them, not knowing if you know what they are.

let $a$ be the side of the basis and $\theta$ the angle you're looking for. Then the surface $S$ of the triangular side scales as $a^2/\cos \theta$ (I mean that $S=c a^2/\cos\theta$ for some constant $c$ not of interest).The volume $V$ scales as $a^3 \tan \theta$.

So $S\sim a^2/\cos\theta$ and $V\sim a^3\tan\theta$. To maximise $V$ is the same as maximising $V^2$, since it is a positive quantity, then you have

$$V^2 \sim a^6(\tan\theta)^2 \sim S^3 (\sin\theta)^2\cos\theta\,.$$

Since $S$ is fixed all you have to do now is to find the $\theta$ maximising $(\sin\theta)^2\cos\theta$, for $\theta\in[0,\pi/2]$. Can you conclude now?

Answer 2

If you want to use Lagrange Multiplier

Pyramid height h; pyramid base a ; $x=h/a$ ;

$$ A = a^2\sqrt{4x^2 +1},\quad V= a^3 x/3\; ;$$ $$\dfrac{\partial A/{\partial a}}{\partial A/{\partial x}}= \dfrac{\partial V/{\partial a}}{\partial V/{\partial x}}$$ Avoiding algebraic calculations $$ 2 x^2=1\; x=\dfrac{1}{\sqrt 2}$$ $$\tan \alpha =\dfrac{2h}{a}\rightarrow 54.75^o$$