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I'm trying to use Pythagoras. Assuming $ a=b, v = 2a + c $ I tried calculating height (Vc) on c. Vc by expressing it with a & c. And then using one of the variables a or c in a function to calculate the plane area of the triangle and then looking up the extremes.

But I'm completely confused here. How can I approach solving this?

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    Does "Pitagor" mean [Pythagoras](http://en.wikipedia.org/wiki/Pythagoras)?2012-12-30
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    Yes sorry ill correct it2012-12-30
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    I'm lost: what does it mean "a triangle with biggest plane are...*and with volume one*? If it is an euclidean triangle it is plane, so what volume and of what is that??2012-12-30
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    I used translate and it tanslated to volume, what I'm looking for would be Scope ? Or combined length of all 3 sides ? I dont know how to expres this.2012-12-30
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    Perimeter? From what language are you translating?2012-12-30
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    Yes Perimeter thats it.2012-12-30
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    Also, how much inequalities do you know? Do you know calculus?2012-12-30
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    I know decent amout of inequalities, from calculus limits, derivatives.2012-12-30
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    I recommend you use the semi-perimeter (Heron's formula) for the area since you know that $s=(2a + c)/2 = 1/2$ and $A = \sqrt{s(s-a)(s-a)(s-c)}$. A few substitutions and you should have a straight forward equation to maximize.2012-12-30
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    As the height to the base in an isosceles triangle is also the median to the base, the height's length is, by Pythagoras Theorem, $$\sqrt{a^2-\left(\frac{c}{2}\right)^2}$$with a=sides' length, c= base's length2012-12-30

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If the length of the base is $b$ the two equal sides must be $\frac{1-b}{2}$ and the altitude is $\sqrt{\left(\frac{1-b}{2}\right)^2-\left(\frac{b}{2}\right)^2}$. Thus, the square of the area is $$ \begin{align} \left(\frac{b}{2}\right)^2\left(\left(\frac{1-b}{2}\right)^2-\left(\frac{b}{2}\right)^2\right) &=\frac{b^2}{4}\left(\frac14-\frac b2\right)\\ &=\frac{b^2}{16}-\frac{b^3}{8} \end{align} $$ Taking the derivative and setting to $0$ yields $$ \frac{b}{8}-\frac{3b^2}{8}=0 $$ which gives $b=0$ or $b=\frac13$. Thus, we get $b=\frac13$ and the triangle is equilateral.

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    Another way to end (just for fun) is that $\frac{1}{16}(b^2)(1-2b)\le \frac{(b+b+(1-2b))^3}{432}=\frac{1}{432}$ with equality iff $b=b=1-2b$ or $b=\frac{1}{3}$.2013-01-02
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    @Apple: indeed. A nice use of the AM-GM inequality. Why not expand this into an answer?2013-01-02
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First write the expression for the area of an isosceles triangle.

$ A(a,b)=\frac{1}{4b}\sqrt{4a^{2}-b^{2}} $

Then use the constraint that $2a+b=1$. This will reduce your area function to one variable $A(a)$. You can then set $A^{\prime}(a)=0$ and find the maximum area.

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Using the semi-perimeter formula which states that $A=\sqrt{s(s-a)^2(s-c)}$, where $s=\frac{1}{2}(2a+c)$. We know that $2a+c=1$, therefore $c=1-2a$ and $s=\frac{1}{2}$. Substituting into the area formula we get

$$A=\sqrt{\frac{1}{2}(\frac{1}{2}-a)^2(\frac{1}{2}-c)}=\sqrt{\frac{1}{2}(\frac{1}{2}-a)^2(-\frac{1}{2}+2a)}$$

First notice that the area is zero for $a=1/4$ (i.e. $c=1/2$, zero height) and $a=1/2$ (i.e. $c=0$, zero width). Geometrically we cannot have $a<1/4$ because the it would violate the triangle inequality. Also, geometrically we cannot have $a> 1/2$ because we could not have a perimeter of 1 (i.e. $c$ would have to be negative). So the maximum clearly lies between the following bounds

$$ 1/4 < a < 1/2.$$

To find the maximum (without loss of generality) we can look for extremes of the following:

$$A^2=\frac{1}{2}(\frac{1}{2}-a)^2(-\frac{1}{2}+2a)$$

Taking the derivative and setting to zero yields the following after simplification:

$$(\frac{1}{2}-a)(1-3a)=0.$$

Which yields a maximum for $a=1/3$ (i.e. $c=1/3$). The maximum area for a fixed perimeter is an equilateral triangle, we might have guessed as much.

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    I like that you took the derivative of $A^2$, which was easier to deal with just from the lack of square roots. However, Your "first notice that..." should come before A^2, as that gives you the bounds that you're working in. Otherwise, the maximum of $A^2$ will be at $\infty$, because you didn't check the end points.2012-12-30
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    @CalvinLin, Thank you for your suggestion. I made the recommended edit.2012-12-30