A man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is a further 20km down the shore from A. If he can row at 8km/hr and run at 10km/r, how far from A should he land?
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max and min problem (1 Viewer)
- Thread starter bos1234
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ssglain
Member
I am a bit surprised that this post still hasn't been moved. Usually it takes about 5 seconds for the mods to do something about it.
Nonetheless.
I'm assumimg the question meant that perpendicular distance AP = 6 because it would otherwise be impossible to work out an answer seeing as no angle is given.
Construct a right-angled triangle ABP, with angle(PAB) = 90.
Suppose the man lands at point M on the side AB, x km away from A.
Then walking distance = MB = (20 - x) km & rowing distance = MP = √(36 - x²) km
Clearly the time taken to travel the route from P to M to B is
T = [√(36 - x²)]/8 + (20 - x)/10
Differentiation gives dT/dx = x/[8√(36 - x²)] - 1/10
Solve for stat pts by setting dT/dx = 0. This gives the minimum T at x = 8 km.
Nonetheless.
I'm assumimg the question meant that perpendicular distance AP = 6 because it would otherwise be impossible to work out an answer seeing as no angle is given.
Construct a right-angled triangle ABP, with angle(PAB) = 90.
Suppose the man lands at point M on the side AB, x km away from A.
Then walking distance = MB = (20 - x) km & rowing distance = MP = √(36 - x²) km
Clearly the time taken to travel the route from P to M to B is
T = [√(36 - x²)]/8 + (20 - x)/10
Differentiation gives dT/dx = x/[8√(36 - x²)] - 1/10
Solve for stat pts by setting dT/dx = 0. This gives the minimum T at x = 8 km.