calculus - finding the maximum area of 2 circles

An equilateral triangle with height $h$ has 2 different incircles.

the bottom circle is tangent to the base of the triangle at the middle point of the base.

what should be the radius of the upper circle so the sum of the area of the circles will be maximum?

i tried to to find connection between the radius and $h$ but i got that the radius should be $0$ and it doesn't seem right.

thanks.

btw, the answer should be $\frac{1}{9}h$

Answer

If the lower circle is inscribed as above (with radius $DE=r$) then, the maximum circle that can be inscribed above it is the inscribed circle having the three sides of $\Delta CGH$ as tangents, $GH$ parallel with $AB$ and $CE=h$ perpendicular to $AB$. which implies that $\Delta CGH$ is also equilateral. With similar triangles one can work out the side length of $\Delta CGH$ which is $\cfrac {2\sqrt 3(h-2r)}{3}$ and the radius of its inscribed circle is $\cfrac {\sqrt 3}{6}\cdot \cfrac {2\sqrt 3(h-2r)}{3}$ and the only way this can be equal to $\cfrac h9$ is if $r=\cfrac h3$

UPDATE: We have two circles of radii, $r$ and $\cfrac{h-2r}3$ and the sum of their areas is $\pi r^2+\pi \cfrac{(h-2r)^2}9$. You can reduce the problem to finding the value of $r$ for which this area is maximum and then substitute the value in the expression above.

$$\max_ {0\le r \le h/3}\left( \pi r^2+\pi \cfrac{(h-2r)^2}9\right)$$