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Math Help - Application of maxima and minima

  1. #1
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    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
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by ticbol View Post
    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
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  4. #4
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    Quote Originally Posted by ticbol View Post
    -----------------------------------------------
    "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.
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  5. #5
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    Quote Originally Posted by ticbol View Post
    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.
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