Brandon is on one side of a river that is 50 m wide and wants to reach a point 200 m downstream on the opposite side as quickly as possible by swimming diagonally across the river and then running the rest of the way. find the minimum amount of time if brandon can swim at 1.5 m/s and run at 5 m/s. (round your answer to two decimal places.)

71.8 seconds

Step-by-step explanation:

In the diagram, the plot if the situation is shown. Brandon wants to go from A to C swimming at 1.5 m/s and from C to D running at 5 m/s.

From speed definition:

time = distance/speed

From pythagorean theorem

AC = √(x² + 50²)

Then the distance AC is done in:

time = √(x² + 50²)/1.5       (in seconds)

On the other hand,  the distance CD is covered in:

time = (200 - x)/5        (in seconds)

The total time is

f(x) = √(x² + 50²)/1.5 + (200 - x)/5

We want to optimize it, then we need to find its first derivative and equalize it to zero:

f(x) = √(x² + 50²)/1.5 + (200 - x)/5

f(x) = √(x² + 50²)/1.5 + 40 - x/5

f'(x) =  x/[1.5*√(x² + 50²)] - 1/5 = 0

x/[1.5*√(x² + 50²)] = 1/5

5*x = 1.5*√(x² + 50²)

5²*x² = 1.5²*(x² + 50²)

25*x² - 2.25*x² = 1.5²*50²

22.75*x² = 5625

x = √(5625/22.75)    

x = 15.72

(the negative result is not taking into account because that solution of the square root doesn't have physical sense for the problem)

Then the minimum amount of time is:

f(15.72) = √(15.72² + 50²)/1.5 + (200 - 15.72)/5 = 71.8 seconds

false

step-by-step explanation:

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.