Calculus I Applied Optimization Problem- How far along the shore should Conner run before jumping into the water in order to save the child?

Question

Conner is a lifeguard and spots a drowning child 60 meters along the shore and 70 meters from the shore to the child. Conner runs along the shore for a while and then jumps into the water and swims from there directly to the child. Conner can run at a rate of 4 meters per second and swim at a rate of 0.9 meters per second. How far along the shore should Conner run before jumping into the water in order to save the child? Round your answer to three decimal places.

Answer 1

From this problem, we need to find the optimal distance that Connor will run and swim that will give the least amount time to get to the drowning child.

First we need to consider how long Connor will be running. If we consider x to be the spot along the shore he stops running, he runs for 60-x meters at 4 m/s. Thus (60-x)/4 seconds is spent running. Now, we need to see how long he will be swimming.

If x, the part of the shore he did not run, is leg a and 70 meters up to the child is leg b, connecting a line c to the child will be the hypotenuse. This is the route he will swim. We calculate the length using the pythagorean theorem.

x² + 70² = c²

c = √ x² + 70² length Connor will swim

To find the seconds he spends swimming, divide by 0.09 meters per second. So he spends (√ x² + 70²)/0.9 seconds swimming. So if we add the time he spends running and the time he spends swimming, we get a function

f(x) = (60-x)/4 + (√ (x² + 70²))/0.9

And remember, we can take the derivative of a function to find the min using critical points. So, if we find f'(x), we'll get an x value (shoreline that Connor didn't run) that will give us the shortest amount of time spent to get to the drowning child.

f'(x) = [10x/ 9√(x²+4900)]− 1/4

We set f'(x) = 0 to find our critical points, which gives us

x = 16.164 meters that Connor didn't run

So Connor should run 60 - 16.164 = 43.836 meters before swimming to the child.