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Math Help - optimization plz help!!

  1. #1
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    optimization plz help!!

    A man can row at 6 km/h, and run at 8 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 10 km, and the water is 6 km wide. He starts rowing with an angle between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of angle

    Thank you so much!!
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  2. #2
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    Quote Originally Posted by inhae772 View Post
    A man can row at 6 km/h, and run at 8 km/h. He needs to get from a point A, on the south bank of a stretch of still water, to point B on the north bank of the water. The direct distance from A to B is 10 km, and the water is 6 km wide. He starts rowing with an angle between North and the direction in which he rows. Find an expression for the time he will take to get from A to B, in terms of angle

    Thank you so much!!
    1. From the definition of speed: speed=\dfrac{distance}{time} you know that

    time = \dfrac{distance}{speed}

    2. From the headline of your post I believe that you want to know the shortest time to go from A to B(?).

    3. Draw a sketch. Actually you are dealing with a right triangle with the length of the hypotenuse of 10 km and the legs, which are 6 km and 8 km long.

    4. Let \theta denote the angle with respect to North (=heading). Then the first part of the trip has a length of:

    d_1=\dfrac6{\cos(\theta)} and the time which takes this part is:  t_1=\dfrac{d_1}{6\ \frac{km}{h}} = \dfrac1{\cos(\theta)}

    5. The second part is:
    d_2=8-6\cdot \tan(\theta) and the time which takes this part is:  t_1=\dfrac{d_2}{8\ \frac{km}{h}} = 1-\dfrac34\cdot\tan(\theta)

    6. The total time is:

    t(\theta)= \dfrac1{\cos(\theta)} +  1-\dfrac34\cdot\tan(\theta)

    7. If you want to get the minimum of the elapsed time, differentiate t, set t(\theta) = 0 and solve for \theta.
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    Thank you so much
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