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Math Help - Optimization: maximize a triangle surface

  1. #1
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    Optimization: maximize a triangle surface

    See if anyone can help me with this: Among all triangles of perimeter equal to P, find the one with the largest area. (Hint: use the formula A=\sqrt[ ]{p(p-x)(p-y)(p-z)} where P=2p, P is the perimeter).

    So, I have f|_s , I think that must be solved using Lagrange multipliers, at least I don't see any other way.

    I've proceeded this way: f=\sqrt[ ]{p(p-x)(p-y)(p-z)}, s=p=\displaystyle\frac{x+y+z}{2}

    Well, I have done so, but all derivatives did wrong (I did A^2 arising as if f=A^2 and then apply the multiplier to with the 4 conditions ), it became ugly, maybe it was because of that. Anyway, would you tell me if what I did until here is ok? Greetings.
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  2. #2
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    First, since \sqrt{x} is an increasing function, any maximum value of \sqrt{x} is a maximum value of x so it is sufficient to maximize A^2= p(p-x)(p- y)(p- z). Taking, as constraint, x+ y+ z= 2p, a constant, you want to maximize F(x, y, z)= p(p- x)(p- y)(p- z) with constraint G(x, y, z)= x+ y+ z= 2p. You want to find x, y, z such that \nabla F= \lambda\nabla G for some constant \lambda. Don't forget that p is a constant.
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  3. #3
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    Lets see then if what I did its ok:

    A^2=p(p^2-py-xp+xy)(p-z)=p^4-p^3y-p^3x+p^2xy-p^3z+p^2zy+p^2zx-pzxy
    Then I call: f=p(p^2-py-xp+xy)(p-z)=p^4-p^3y-p^3x+p^2xy-p^3z+p^2zy+p^2zx-pzxy

    And s=p=\displaystyle\frac{P}{2}=\displaystyle\frac{x+  y+z}{2}

    \begin{Bmatrix}f_x=p(py+pz-zy-p^2)\\f_y=p(px+pz-zx-p^2)\\f_z=p(px+py-p^2-xy)\end{matrix}

    p\neq{0}
    Then:
    \begin{Bmatrix}py+pz-zy-p^2=\displaystyle\frac{\lambda}{2}\\px+pz-zx-p^2=\displaystyle\frac{\lambda}{2}\\px+py-p^2-xy=\displaystyle\frac{\lambda}{2}\\p=\displaystyle  \frac{x+y+z}{2} \end{matrix}

    From the last condition I replaced p to get lambda.
    Then I've found that:
    \displaystyle\frac{\lambda}{2}=-x^2+y^2+z^2-2zy
    Then I replace in one of the equations

    -x^2+y^2+z^2-2zy=\displaystyle\frac{(x+y+z)}{2}x+\displaystyle\  frac{(x+y+z)}{2}z-zx-(\displaystyle\frac{(x+y+z)}{2})^2
    So I get into what I think its an absurd:
    z=\displaystyle\frac{2y^2-2x^2}{2y-2x}
    I think its an absurd, but I haven't go ahead from hear. And the reason is that I think I should have an isosceles triangel, so I would be dividing be zero on this equality.
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