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Math Help - Optimization

  1. #1
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    Optimization

    For a given perimeter, what type of triangle encloses the most area.

    How do I do this since triangles have three sides, a b and c for perimeter?
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  2. #2
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    Quote Originally Posted by millerst View Post

    For a given perimeter, what type of triangle encloses the most area.

    How do I do this since triangles have three sides, a b and c for perimeter?
    there are so many ways to solve this problem. i'll give you two of them. i'll assume that x,y,z are the length of the sides and x+y+z=p, where p is fixed.

    Solution 1: by Heron's formula: S(x,y,z)=\frac{\sqrt{p}}{4} \sqrt{(p-2x)(p-2y)(p-2z)}=\frac{\sqrt{p}}{4} \sqrt{(p-2x)(p-2y)(2x+2y-p)}. so we only need to maximize the function

    f(x,y)=(p-2x)(p-2y)(2x+2y-p). so we must solve theis system: \frac{\partial{f}}{\partial x}=\frac{\partial{f}}{\partial x}=0, which will easy give you x=y=\frac{p}{3}, and thus z=\frac{p}{3}. so the equilateral

    triangle with the side of length \frac{p}{3} is the answer.

    Solution 2: following our notation in solution 1, by AM-GM inequality we have: \frac{p}{3}=\frac{p-2x + p-2y + p-2z}{3} \geq \sqrt[3]{(p-2x)(p-2y)(p-2z)}, which gives us:

    S(x,y,z) \leq \frac{p^2}{12 \sqrt{3}}=S \left(\frac{p}{3}, \frac{p}{3}, \frac{p}{3} \right). \ \Box
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  3. #3
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    Hello, millerst!

    This problem requires familiarity with partial derivatives.


    For a given perimeter, what type of triangle encloses the most area?
    Let the three sides be: . x,y,z

    The perimeter is a constant, P\!:\;\;x + y + z \:=\:P \quad\Rightarrow\quad z \:=\:P - x - y


    We can use Heron's Formula: . \text{Area} \:=\:\sqrt{s(s-x)(s-y)(s-z)}

    . . where s is the semiperimeter: . s \:=\:\frac{x+y+z}{2} \:=\:\frac{P}{2}


    To maximize the area, we can maximize the square of the area.

    That is, use the square of Heron's Formula: . A \;=\;s(s-x)(s-y)(s-z)\;\;{\color{blue}[1]}


    We find that: . \begin{array}{ccc}s-x &=&\dfrac{P-2x}{2} \\ \\[-4mm]s-y &=& \dfrac{P-2y}{2} \\ \\[-4mm] s-z &=& \dfrac{2x+2y-P}{2} \end{array}


    Then [1] becomes: . A \;=\;\frac{P}{2}\left(\frac{P-2x}{2}\right)\left(\frac{P-2y}{2}\right)\left( \frac{2x+2y-P}{2}\right)

    which simplifies to: . A \;=\;\frac{P}{16}\left(4P^2x + 4P^2y - 4Px^2 - 4Py^2 - 12Pxy + 8x^2y + 8xy^2 - P^3\right)


    Equate the partial derivatives to zero:

    . . \begin{array}{ccccccc}\dfrac{\partial A}{\partial x} &=& \frac{P}{16}\left(4P^2 - 8Px - 12Py +16xy + 8y^2\right) &=& 0 & {\color{blue}[2]}\\ \\[-3mm]<br />
\dfrac{\partial A}{\partial y} &=& \frac{P}{16}\left(4P^2 - 8Py - 12Px + 16xy + 8x^2\right) &=& 0 & {\color{blue}[3]}\end{array}


    We have: . \begin{array}{cccc}P^2 - 2Px - 3Py + 4xy + 2y^2 &=& 0 & {\color{blue}[4]}\\ \\[-3mm]<br />
P^2 - 2Py - 3Px + 4xy + 2x^2 &=& 0 & {\color{blue}[5]}\end{array}


    Solve [5] for y\!:\;\;2Py - 4xy \:=\:p^2-3Px+2x^2 \quad\Rightarrow\quad2(P-2x)y \:=\:(P-x)(P-2x)

    . . y \:=\:\frac{(P-2)(P-2x)}{2(P-2x)} \quad\Rightarrow\quad y\:=\:\frac{P-x}{2} .**


    Substitute into [4]: . P^2 - 2Px - 3P\left(\frac{P-x}{2}\right) + 4x\left(\frac{P-x}{2}\right) + \left(\frac{P-x}{2}\right)^2 \:=\:0

    . . which simplifies to: . \tfrac{1}{2}Px - \tfrac{3}{2}x^2\:=\:0 \quad\Rightarrow\quad \boxed{x \:=\:\tfrac{P}{3}}

    Back-substitute and we get: . \boxed{y \:=\:\tfrac{P}{3}}\;\;\boxed{z\:=\:\tfrac{P}{3}}


    As you probably suspected, the triangle of maximum area is equilateral.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    **

    We can cancel P - 2x.

    If P - 2x \:=\:0, then: . x \:=\:\frac{P}{2}

    If side x is half the perimeter, it violates the Triangle Inequality.



    Edit: Nicely done, NonCommAlg!
    .
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