Sunday, 14 August 2016

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 19h


Answer



enter image description here
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 ΔCGH as tangents, GH parallel with AB and CE=h perpendicular to AB. which implies that ΔCGH is also equilateral. With similar triangles one can work out the side length of ΔCGH which is 23(h2r)3 and the radius of its inscribed circle is 3623(h2r)3 and the only way this can be equal to h9 is if r=h3



UPDATE: We have two circles of radii, r and h2r3 and the sum of their areas is πr2+π(h2r)29. 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.

max0rh/3(πr2+π(h2r)29)


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