Juan is a lifeguard and spots a drowning child 50 meters along the shore and 40 meters from the shore to the child.

Question

Juan is a lifeguard and spots a drowning child 50 meters along the shore and 40 meters from the shore to the child. Juan runs along the shore for a while and then jumps into the water and swims from there directly to the child. Juan can run at a rate of 4 meters per second and swim at a rate of 1.1 meters per second. How far along the shore should Juan run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

Child at point C, lifeguard at A.

Point B is 50 meters to the left of A and 40 meters below C.

Right triangle ABC with legs 50 and 40. The lifeguard is

sqrt( 4100) away from child

Let x be the running distance and point P be the point

where the lifeguard dives into the water.

Then AP = x and BP = 50-x

Then right triangle CBP has legs 40 and 50-x

The swimming distance is then

sqrt( 40^2 + (50-x)^2 ) =

sqrt( 1600 + 2500 - 100x + x^2)

= sqrt( x^2 - 100x + 4100)

The total distance function is then:

D(x) = x + sqrt( x^2 - 100x + 4100)

Minimizing:

0 = dD/dx = 1 + (2x-100) / sqrt( x^2 - 100x + 4100)

- sqrt( x^2 - 100x + 4100) = 2x-100

squaring both sides:

x^2 - 100x + 4100 = 4(x-50)^2

x^2 - 100x + 4100 = 4 x^2 - 400x + 10000

0 = 3x^2 - 300x + 5900

quadratic formula results in x = 73.094 and x = 26.906

At 4 meters per second, the life guard runs for 6.7265 seconds

before diving into the water. He will then swim for 42 seconds

before reaching the child.