Khushi S.

asked • 04/06/22

Could someone help me with this max/min word problem?

A messenger is to go ashore from a ship 13 km offshore and deliver a message to a camp which is 32 km up the beach from the point nearest the ship. He can go 24 km/h by boat and will be met by a jeep which can travel 40km/h over the beach. Where should he land to compete the trip in the shortest possible time?


(answer is 22.25 km)


thanks so much.

1 Expert Answer

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Marion J. answered • 04/06/22

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Using this sketch, you can create a function for the time it will take the messenger to reach the camp.

Remember that Time = Distance/Speed


T = √(x2 + 132)/24 + (32 - x)/40

= 1/24(x2 + 169)1/2 + 4/5 - x/40


Differentiate, put the derivative equal to zero, and solve for x:

x/24(x2 + 169)-1/2 - 1/40 = 0

x = 9.75


Therefore, the messenger must land 32 - 9.75 = 22.25 km from the camp.

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Khushi S.

when you went from T = √(x^2 + 132)/24 + (32 - x)/40 to 1/24(x^2 + 169)^1/2 + 4/5 - x/40 how did you get the 1 in the numerator and 24(x^2 + 169)^1/2 in the denominator? was that simplifying or were you getting the derivative of that part?
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04/08/22

Khushi S.

also how did you get a negative exponent (-1/2) in the derivative?
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04/08/22

Marion J.

The square root over 24 can be rewritten as 1/24 × the square root. The square root becomes an exponent of 1/2. When you differentiate, subtracting 1 from the exponent gives - 1/2.
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04/08/22

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