1. ## 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!!

2. Originally Posted by inhae772
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: $\displaystyle speed=\dfrac{distance}{time}$ you know that

$\displaystyle time = \dfrac{distance}{speed}$

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

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

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

6. The total time is:

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

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

3. Thank you so much