Find a isosceles triangle with biggest plane area with perimeter 1

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? – Zev Chonoles Dec 30 '12 at 3:32
Yes sorry ill correct it – Sterling Duchess Dec 30 '12 at 3:35
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?? – DonAntonio Dec 30 '12 at 3:37
Perimeter? From what language are you translating? – DonAntonio Dec 30 '12 at 3:41
Also, how much inequalities do you know? Do you know calculus? – Calvin Lin Dec 30 '12 at 3:47

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 $b=\frac{1}{3}$. – Apple Jan 2 '13 at 18:37
@Apple: indeed. A nice use of the AM-GM inequality. Why not expand this into an answer? – robjohn Jan 2 '13 at 18:46

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. – Calvin Lin Dec 30 '12 at 6:01
@CalvinLin, Thank you for your suggestion. I made the recommended edit. – Tpofofn Dec 30 '12 at 12:40

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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