Maximisation: Half circle area inscribed within isoceles triangle? Given an isosceles triangle with the line of symmetry along $x=0$, and the odd side along $y=0$, How can I optimize the maximum parabolic area inscribed within the triangle? The "half -circular" area has points of tangency to the triangle, one of each side of the line of symmetry. What I am having trouble is, I simply don't know how to start, since I don't know the dimensions of of the triangle which the problems usually give. I have tried looking only at its "first quadrant", trying to find a negative parabola tangent to the line (triangle side) with a negative slope, but I still can't figure it out. Thanks in advance!
 A: Let the triangle have side with equation $y_1=c-mx$. Start here.
Let the parabola have equation $y_2=b-ax^2$.
Gradients are $\dfrac{dy_1}{dx}=-m$ and $\dfrac{dy_2}{dx}=-2ax$
Gradients equal when $m=2ax$, this is ${x}=\dfrac{m}{2a}$.
Must also have same $y$ value, so
$$c-m\frac{m}{2a}=b-a\frac{m}{2a}\frac{m}{2a}$$
$$c-\frac{m^2}{2a}=b-\frac{m^2}{4a}$$
$$4ac-2m^2=4ab-m^2$$
$$4a(c-b)=m^2.$$
$$a=\frac{m^2}{4(c-b)}$$
Area under curve is $\int_{-\sqrt{\frac{b}{a}}}^{\sqrt{\frac{b}{a}}} {b-ax^2}dx$
$$A_{curve}=\frac{4}{3} \sqrt{\frac{b^3}{a}}$$
$$A_{triangle}=\frac{\frac{2c}{m} c}{2}=\frac{c^2}{m}$$
$$\frac{A_{curve}}{A_{triangle}}=\frac{4m}{3c^2} \sqrt{\frac{b^3}{a}}$$
$$=\frac{4m}{3c^2} \sqrt{\frac{4b^3(c-b)}{m^2}}$$
$$=\frac{8}{3c^2} \sqrt{b^3(c-b)}$$
Differentiate this wrt b
$$\frac{d{Area}_{proportion}}{db}=\frac 1 2 {(cb^3-b^4)}^{\frac{-1}{2}} (3cb^2-4b^3)$$
Minimums means $ (3cb^2-4b^3)=0$
$$3c=4b$$
$$b=\frac {3c}{4}$$
Substituting gives $\frac{A_{curve}}{A_{triangle}}=\frac{\sqrt{3}}{2}$
A: Having contributed one answer, I felt I could simplify this!
If we consider the triangle formed by the lines $y=1-x$ and $y=1-x$, the problem is easier.
But  the answer will be the same as any other triangle will be an enlargement of this triangle.
