OL Exam Question on Relative velocity

[pcpssam]

correction -->it should be a high-level exam question.

Homework Statement

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 traveling 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 don't understand..how can finding the maxium value
of tanB help me to find the shortest distance?

 

[haruspex]

It's the min of tan B, not the max.
The minimum distance is 30 sin B, so want the value of A that minimises sin B.
d sin(B)/dA = d sin(B)/d tan(B) * d tan(B) /dA (chain rule), so for d sin(B)/dA = 0 we want either d sin(B)/d tan(B) = 0 or d tan(B) /dA = 0. The first never happens.