Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 4 / 16C / c

Here $b>0$ because $a=0$ and the two cost coefficients are not both zero. The interior Euler-Lagrange equation makes $V$ quadratic. Imposing each displayed pair of endpoint data and the distance constraint gives the following formal profiles and their costs:Thus choice (3) is the best strategy, and its true minimum cost isThe qualification from part (b) applies to row (1): its formal quadratic is not a minimizer with free terminal speed, since $\dot{V}(T)\ne 0$. Row (2) is a genuine minimizer for its fixed zero terminal speed, and row (3) is the genuine free-terminal-speed minimizer, satisfying its natural condition $\dot{V}(T)=0$.

A direct lower bound makes the comparison rigorous. For every admissible speed with $V(0)=0$,by integration by parts. The Cauchy-Schwarz inequality gives $L^2\le (T^3/3){\int }_0^T{\dot{V}}^{2}dt$. Equality requires $\dot{V}$ to be proportional to $T-t$, giving exactly row (3). Row (1)'s zero initial acceleration excludes equality; however short initial layers, as in part (b), approach that same infimum. Hence choice (1) has infimum $3b{L}^2/T^3$ but does not attain it, while choice (3) does. This distinction resolves the boundary-condition issue in the printed minimization request.