correction -->it should be a high-level exam question. 1. The problem statement, all variables and given/known data Two ships A and B move with constant speeds 48 km/h and 60 km/h respectively. At a certain instant ship B is 30 km west of A and is travelling due south. Find (i) the direction A should steer in order to get as close as possible to ship B (ii) the shortest distance between the ships. the answer: Diagram 1: Vab = Va -Vb Vab = (-48cosAi - 48sinAj) - (-60j) = (-48cosA)i + (60-48sinA)j tanB = j over i = 60-48sinA over 48cosA tanB = 5-4sinA over 4cosA then differentiate tan B with respect to A and make it equal to 0 You get 16cos^2 A - 20sinA + 16sin^2 A which simplifies to sin A = 4/5 , (and tanB = 3/4) Then part 2: (from diagram 2) |BX| = 30sinB = 30(0.6) = 18km Why I need to differentiate tanB equal to 0? I just dont understand..how can finding the maxium value of tanB help me to find the shortest distance?