Optimization Application Problem

Question

The potential energy U of a particle moving along the x-axis is given by U= b(a^2/x2 - a/x) where a and b are positive constants and x> 0 . What value of x minimizes the potential energy?

Answer 1

U' = b(- 2a²/x³ + a/x²) = 0; ax - 2a² = 0; x = 2a.

To prove that at x = 2a we have minimum let use 2nd derivative test: U'' = b(6a²/x⁴ - 2a/x³).

At x = 2a U'' = b(6a²/16a⁴ - 2a/8a³) = b(3/(8a²) - 2/(8a²)) = b/(8a²) > 0. So, we have minimum at x = 2a.

Minimum of Potential energy: U_min = b(a²/(4a²) - a/(2a)) = - b/4