Question

find the lengths of the sides of an isosceles triangle with a given perimeter if its area is to be s great as possible

Answer 1

In general, maximizing area (or volume) given a perimeter (or area) limit is done by most closely approximating a circle (or sphere). Your triangle should turn out to be equilateral, three equal sides. Maximizing area holding perimeter constant involves using the perimeter to express one of the variables and then using that in the area equation, differentiating it and setting it to zero.

Answer 2

'CAUSE I'M ridiculous and can't do maths.
\[ a^2 = \frac{x^2}{4}+y^2\]

Answer 3

I ended up with P = 4a, which I did not believe. I think x=a, P=3a is the correct result, despite my math.

Answer 4

I did not work with y as a variable, going right to a^2-x^2/4=h^2

Answer 5

\[P=\sqrt{x^2+4y^2}+x\]
\[y^2 = \frac{(P-x)^2 - x^2}{4}\]
\[y=\frac{1}{2}\sqrt{P^2-2Px)}\]

\[A = \frac{1}{2}xy\]
\[=\frac{x}{4}\sqrt{P^2-2Px)}\]

\[ \frac{\mathrm{d}A}{\mathrm{d}x} = \frac{1}{4}\left(\sqrt{P^2-2Px)}-\frac{xP}{\sqrt{P^2-2x)}}\right) \]
\[P^2-3Px=0\]
\[x=\frac{P}{3}\]

And right here we can just read off the answer from the original definition of the perimeter
\[P = 2a + x\]
\[P = 2a + \frac{P}{3}\]
\[a = \frac{P}{3}\]
So. like @douglascooperwinslow said, the maximized area is an equilateral triangle.

For kicks and grins, to find the height of the triangle we can use

\[y=\frac{1}{2}\sqrt{P^2-2Px}\]
\[y=\frac{1}{2}\sqrt{P^2-\frac{2P^2}{3}}\]
\[y=\frac{P \sqrt{3}}{6}=\frac{P}{3} \frac{\sqrt{3}}{2}=a\frac{\sqrt{3}}{2}\]

Answer 6

derp, sorry.
@douglaswinslowcooper

Answer 7

Very well done! So well done, it's rare. Regards.