Can't figure out how to solve this please help! Thank you!

Question

Zoey is a lifeguard and spots a drowning child 60 meters along the shore and 50 meters from the shore to the child. Zoey runs along the shore for a while and then jumps into the water and swims from there directly to the child. Zoey can run at a rate of 3 meters per second and swim at a rate of 0.9 meters per second. How far along the shore should Zoey run before jumping into the water in order to save the child? Round your answer to three decimal places

Answer 1

You need to minimize for time. This is a d=rt problem, so t = time to run along the beach plus time to swim

to the child.

Let x = beach distance and s = the swimming distance.

The distance s is the hypotenuse of a triangle with base of (60-x) and height of 50 meters. If you sketch this,

it will make sense.

s = √(50²+(60-x)²)

Total time = t = f(x) = x/3 + s/.9 = x/3 + √(50²+(60-x)²)/.9

Take the derivative of f(x), set it to zero and solve for x, which is what the problem is wanting.

f '(x) = 1/3 - (1/.9)(2(60-x))/(2√(50²+(60-x)²))=1/3-(60-x)/√(50²+(60-x)²)(.9)=0

x≈44.276 meters

Answer 2

If lifeguard swims the diagonal, it takes 86.8s., If he runs 60m and swims 50m. it takes 75.6s

t=[(60-x)/3]+(50²+x²)^(1/2)/.9

60-x is distance ran

dt/dx=(-1/3)+((1/2(50²+x²)^(-1/2)2x/.9

set to zero and solve for positive x

x=15.7 meters (rounded) meaning lifeguard runs 44.3 meters

time to child is t=73.0s.