I have this problem from Calculus 1 that I need help with. I have no idea where to start.
I would use a bit of symmetry here. Bisect the circle and the isosceles triangle so that you have a right triangle within a semi-circle. Let the semi-circle be described by:
$\displaystyle y=\sqrt{4^2-x^2}$
Let the hypotenuse of the right triangle extend from (-4,0) to (x,y) and the horizontal leg from (-4,0) to (x,0) and the vertical leg from (x,y) to (x,0).
The area A of the triangle is then:
$\displaystyle A=\frac{1}{2}bh=\frac{1}{2}(x+4)(\sqrt{4^2-x^2})$
Now, maximize this function, then double the maximum area to find the maximum area for the original problem.