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Thread: Tricky Word Problem

  1. #1
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    Question Tricky Word Problem

    Hello,

    I have been tasked with solving the problem below and I am really stuck. If anyone can help me figure it out, you rock!

    The Problem:

    Alex the geologist is in the desert, 10 km from a long, straight road and 45 km from base camp. The base camp is also 10 km from the road, on the same side of the road as Alex is. On the road, the jeep can do 50 kph, but in the desert sands, it can manage only 30 kph. Alex wants to return to base camp as soon as possible. How long will the trip take?

    Thanks in advance,
    A desperate student
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  2. #2
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    Re: Tricky Word Problem

    From the symmetry of the problem we know that second half of the trip will be a reflection of the first half of the trip.

    The first half of the trip consists of travelling from his origin at $(0,0)$ to some point on the road, $(x, 10)$, and then continuing to the halfway point on the road at $(22.5,10)$

    The time this takes is

    $T = \dfrac{\|(x,10)\|}{30} + \dfrac{\|(22.5,10)-(x,10)\|}{50}$

    $T = \dfrac{\sqrt{x^2+100}}{30} + \dfrac{\sqrt{(22.5-x)^2}}{50} = \dfrac{\sqrt{x^2+100}}{30} + \dfrac{22.5-x}{50}$

    Minimizing this without calculus will involve a fair bit of algebra.

    Are you allowed to use calculus to find the value of $x$ that minimizes $T$ ?

    If so you simply solve $\dfrac{dT}{dx}=0$ for $x$ and ensure that $\dfrac{d^2T}{{dx}^2}>0$ at this value of $x$

    Doing this you find that $x=\dfrac{15}{2}~km$ and $T = \dfrac{43}{60}~hr = 43~min$
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