1. ## 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?

2. Originally Posted by millerst

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 $\displaystyle x,y,z$ are the length of the sides and $\displaystyle x+y+z=p,$ where $\displaystyle p$ is fixed.

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

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

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

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

$\displaystyle S(x,y,z) \leq \frac{p^2}{12 \sqrt{3}}=S \left(\frac{p}{3}, \frac{p}{3}, \frac{p}{3} \right). \ \Box$

3. 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: .$\displaystyle x,y,z$

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

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

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

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

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

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

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

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

Equate the partial derivatives to zero:

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

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

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

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

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

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

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

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

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**

We can cancel $\displaystyle P - 2x$.

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

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

Edit: Nicely done, NonCommAlg!
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