A unit disk (radius = 1) is partitioned into an inner circular region of radius ( r < 1 ) and an outer region consisting of the area between concentric circles of radii r and 1 respectively.
The inner region is preserved, and the outer region is reformed into a circular disk with equivalent area. Thus, the combined area of the two resulting disks is equal to the area of the original unit disk.
Which value of r (to 3 decimal places) maximizes the sum of the circumferences of the two final disks?
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Relevant wiki: Cauchy-Schwarz Inequality
Let the radius of the inner region be r and the radius of the reformed green area be R .
We are given that their area is equal to the area of the original unit disk, hence
π r 2 + π R 2 = π r 2 + R 2 = 1
We want to maximise the combined circumference 2 π r + 2 π R → 2 π ( r + R ) . Since 2 π is constant, we only have to maximise r + R . By Cauchy-Schwarz Inequality,
( r 2 + R 2 ) ( 1 + 1 ) ≥ ( r + R ) 2 r + R ≤ 2
The condition to satisfy the maximum is when r = R , hence r = 2 2 .
@Steven Chase We think that having MCQ options of exact answers would make this problem more engaging. Would you mind if we made the change?
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Relevant wiki: Second Derivative Test
Here's a calculus approach:
Like what Christopher Boo has written up, we have
r 2 + R 2 = 1 , ( r , R ) ∈ [ 0 , 1 ] × [ 0 , 1 ] .
And we want to maximize f = r + R .
Substituting R = 1 − r 2 gives f = r + 1 − r 2 .
This is a 1-variable calculus problem with f ′ = 1 − 1 − r 2 r .
At the extrema point, we have f ′ = 0 ⇒ r = R = 2 1 .
To prove that it is a local maximum, we apply the second derivative test , to show that f ′ ′ < 0 when r = 2 1 .
f ′ ′ = − ( 1 − r 2 ) 3 / 2 1 < 0 when r = 2 1 .
Since we have established that it is a local maximum, we still need to prove that it's a global maximum. It's trivial to show that at the boundary conditions f ( r , R ) = f ( 0 , 1 ) or f ( 1 , 0 ) yields a lower value than f ( 2 1 , 2 1 ) .
Thus we have proved that the global maximum of the function r + R occurs when r = R = 2 1 . Our answer is r + R = 2 2 = 2 2 .