# Math Help - Application of maxima and minima

1. ## Application of maxima and minima

A pyramid with a square base has a vertex vertically above the centre of its base. The volume of the pyramid is V and the length of each side of the base is x. The total area of the four triangular surfaces is S. Show that $S^2=x^4 + (36V^2 / x^2)$

Show that if V is a constant and x is a variable , then the value of S is the least when $x^3 = (3 sqrt(2)) V$

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

Thanks a lot

2. 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.

------------------------------
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.)

-----------------------------------------------
"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.

3. 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.

------------------------------
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.)

-----------------------------------------------
"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.
Thanks for helping me a lot.
I'm really grateful

4. Originally Posted by ticbol
-----------------------------------------------
"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.
After another approach using variables and still cannot prove it, I will stop my tries now.
Let me use numbers to prove it, if it is true.

Let x = say, 6.

S^2 = x^4 +(36V^2)/(x^2)
S^2 = 6^4 +(36V^2)/(6^2)
S^2 = 1296 +V^2

x^3 = (3sqrt(2))V
V = (6^3) / 3sqrt(2) = 50.91169

So, S^2 = 1296 +(50.91169)^2 = 3888S
And S = 62.35383
And (1/4)S = 15.58846 -------**
That is the area of one triangular face.

If the triangular face is equilateral, with 6 an edge,
area = (1/2)(6)(6)sin(60deg) = 15.58846 -------**

The same! Hey!

Well, sorry, I cannot prove it by variables now. I will leave it now.

5. Originally Posted by ticbol
After another approach using variables and still cannot prove it, I will stop my tries now.
Let me use numbers to prove it, if it is true.

Let x = say, 6.

S^2 = x^4 +(36V^2)/(x^2)
S^2 = 6^4 +(36V^2)/(6^2)
S^2 = 1296 +V^2

x^3 = (3sqrt(2))V
V = (6^3) / 3sqrt(2) = 50.91169

So, S^2 = 1296 +(50.91169)^2 = 3888S
And S = 62.35383
And (1/4)S = 15.58846 -------**
That is the area of one triangular face.

If the triangular face is equilateral, with 6 an edge,
area = (1/2)(6)(6)sin(60deg) = 15.58846 -------**

The same! Hey!

Well, sorry, I cannot prove it by variables now. I will leave it now.
Salute , thanks for everything.