Originally Posted by

**ticbol** The area of one triangular face is (1/2)(x)L

Where L = slant height.

(slant height is not the same as the ridge height.)

So, (1/4)S = (1/2)xL

S = 2xL

S^2 = 4(x^2)(L^2) -----------(i)

L^2 = (x/2)^2 +h^2 -----------(ii)

where h = vertical height of the pyramid

V = (1/3)(x^2)h

So, h = 3V / (x^2)

And so, h^2 = (9V^2)/(x^4) ---------(iii)

Substitute that into (ii),

L^2 = (x^2)/4 +(9V^2)/(x^4) ------(iv)

Substitute that into (i),

s^2 = 4(x^2)[(x^2)/4 +(9V^2)/(x^4)]

S^2 = x^4 +(36V^2)/(x^2) ---------------shown.

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Differentiate both sides of the equation with respect to x,

2S*(dS/dx) = 4x^3 +(36V^2)[-2/ (x^3)]

Set ds/dx to xero,

0 = 4x^3 -(72V^2)/(x^3)

0 = x^3 -(18V^2)/(x^3)

0 = x^6 -18V^2

18V^2 = x^6

(sqrt(18))V = x^3

x^3 = (3sqrt(2))V -----when S is minimum, shown.

(Of course it is not yet shown that that is for minimum S, but you can verify that with further calculation if you like.)

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"Show that in this case, each triangular surface is an equilateral triangle."

Now, here is the the most time-consuming part. I wasted a lot of time for this last part, yet, I could not prove it. Three approaches and I failed in showing it. I am tempted to post one of my calculations to show that at x^3 = (3sqrt(2))V, the triangular face is not equilateral.

Either that one face is not an equilateral triangle, or my mind could not catch the right approach.

So, for now, I leave that last part to you.

If I could solve it later, I'd show it here.