An isosceles triangle with perimeter $12$ is constructed.

1. Find the largest possible area of the triangle.

2. Explain the significance of your solution.

It would be convenient to let the two equal sides have length $a$, and then the base of the triangle have length $2b$ for some $b$, as below:

Because the perimeter is $12$, we have

Also, by Pythagoras, we have

If you want to maximise

it would be sufficient to maximise $A^2$. This makes the problem much easier.

Get $A^2$ in terms of $b$ only, and solve

to find the turning point.

As a fully simplified surd, the area is

Give the value of $q^p$.