Optimization and related rates problem

Question

Ren is a lifeguard and spots a drowning child 50 meters along the shore and 70 meters from the shore to the child. Ren runs along the shore for a while and then jumps into the water and swims from there directly to the child. Ren can run at a rate of 5 meters per second and swim at a rate of 1 meter per second. How far along the shore should Ren run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

Suppose he runs x m along the shore then swims.

Running distance x

Swimming distance : √ 70² + (50-x)²

Time to reach the child = distance/speed = (x/5) + (√(70² + (50-x)²) = (x/5) + √ (7400 + x² - 100x)

Lets call this y

Then we want to minimise y

so we need to find x so dy/dx = 0

dy/dx = (1/5) + 0.5 (7400 + x² - 100x)^-1/2 (2x-100) using chain rule

if this is zero then

-1/5 = (x-50)/ (7400 + x² -100x)^1/2

1/5 = (50-x) / (7400 + x² -100x)^1/2

5(50-x) = (7400 + x² - 100x)^1/2

25 (50-x)² = 7400 + x² - 100x

25 (2500 - 100x + x²) = 7400 + x² - 100x

Giving x = 35.7, 64.3 (I plotted both on Desmos and found points of intersection)

Technically you should check the second derivative is positive to find which one is the minimum (you can do this on a TI-84), giving x = 35.7. Intuitively it's also obvious that x=35.7 has to be the solution as 35.7 < 50