# 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 $\frac{1}{9}h$

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are the circles non intersecting, non overlapping? –  user31280 Jun 8 '13 at 15:44
what is the radius of the lower circle –  user31280 Jun 8 '13 at 15:44
they are Tangential circles and there is no other detail about the radius. but that's just in the picture i got. i don't think that i should rely on it –  bori12 Jun 8 '13 at 15:46
Your calculation may be right. If $x$ is the radius of the lower circle, and $y$ the radius of the upper circle, we find that $3y+2x=h$. Look at the sum of the areas. It has an endpoint maximum. The maximum area is $\pi h^2/9$. –  André Nicolas Jun 8 '13 at 15:56
that's my problem. $\pi h^2/9$ it's the limit i got while $y$ Approaching $\frac{h}{3}$ from the left. so i got zero for the radius –  bori12 Jun 8 '13 at 16:03

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. $$\max_ {0\le r \le h/3}\left( \pi r^2+\pi \cfrac{(h-2r)^2}9\right)$$
Then again, how could i get $r=\frac{h}{3}$ without knowing the answer in advance? and what about the case that $x,y=0$ as mentioned? thanks so much for the answer. –  bori12 Jun 8 '13 at 16:37
so in $\max_ {0\le r \le h/3}\left( \pi r^2+\pi \cfrac{(h-2r)^2}9\right)$ i put: $r=0$ and get $\pi h^2/9$ then i put: $r=\frac{h}{3}$ and get: $10\pi h^2/81$ so the max is for $r=\frac{h}{3}$? –  bori12 Jun 8 '13 at 17:18
yes. and $10\pi h^2/81$ is bigger. another thing, $r$ is the not the radius asked in the question they ask about the upper one. sorry for all the questions. –  bori12 Jun 8 '13 at 17:24
first Derivative and putting the values of the endpoints in the function to see the absolute value for $f$. can you show how so i can learn? –  bori12 Jun 8 '13 at 17:37
so i got $r=\frac{2h}{13}$ min. and $r=0$, $r=\frac{h}{3}$ max.ep and for the endpoints i substitute as i said before. –  bori12 Jun 8 '13 at 18:14