52nd Putnam 1991

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Problem A5

Let f(z) = ∫0z   √(x4 + (z - z2)2) dx. Find the maximum value of f(z) in the range 0 ≤ z ≤ 1.

 

Solution

Moderately hard.

Answer: 1/3.

The integrand is strictly positive for 0 < x ≤ 1 and 0 ≤ z ≤ 1, so f(z) > 0 for 0 < z ≤ 1. However, it is not at all clear how to get the integral in closed form except for z = 0 or 1. So we have a strong suspicion that the maximum must be the higher of the two values we can actually calculate: f(1) = 1/3! To prove this we need to show that f '(z) > 0 for z > 0.

f '(z) = √(z4 + z2(1 - z)2) + z(1 - z)(1 - 2z) ∫0z (x4 + z2(1 - z)2)-1/2 dx. The integrand is positive, so clearly f '(z) > 0 for 0 < z ≤ 1/2. The integrand is less than 1/√(z2(1 - z)2) = 1/z 1/(1 - z), so the integral is less than 1/(1 - z), and for z > 1/2, f '(z) > √(z4 + z2(1 - z)2) - z(2z - 1). But, squaring, we see that z(2z - 1) < √(z4 + z2(1 - z)2) for 1/2 < z < 1, which completes the proof.

 


 

52nd Putnam 1991

© John Scholes
jscholes@kalva.demon.co.uk
21 Sep 1999