(Highschool) AP Calculus, Optimization

Question

Megan is a lifeguard and spots a drowning child 60 meters along the shore and 40 meters from the shore to the child. Megan runs along the shore for a while and then jumps into the water and swims from there directly to the child. Megan can run at a rate of 3 meters per second and swim at a rate of 1.2 meters per second.

How far along the shore should Megan run before jumping into the water in order to save the child? 

Round your answer to three decimal places.

Answer 1

time T = distance divided by speed, d=rt, t=d/r

T= (60-x)/3 +sqr(40^2 +x^2)/1.2

= 20-x/3 +(1/1.2)sqr(1600+x^2)

T'(x)= -1/3 +(1/1.2)(1/2)(2x)/sqr(1600+x^2)=0

-1/3 +x/1.2sq(1600+x^2)=0

x/1.2sqr(1600+x^2)=1/3

x = 1.2sqr(1600+x^2)/3

x^2 = 1.44(1600+x^2)/9

(1-1.44/9)x^2 = 144(16)/9

7.56x^2/9 = 144(16)/(9)

7.56x^2 =144(16)

x^2= 144(16)/7.56=304.7619

x= 24/sqr7.56= 8.72872

run60-x = 51.27128 meters

about 51.271 meters run along the shore

swim sqr(1600+8.729^2) meters

time = 51.27128/3 +sqr(1600+8.729^2/1.2

= 17.09043 +34.118

= about 51.208 seconds

might be errors in above calculations, but checking surronding possibilities shows it's in the right range

20+40/1.2= 20+33 1/3= 53 1/3 seconds if she ran 60 then swam 40

or

she swam sqr(60^2+40^2) then about 60.09 seconds

run 30 then swim 50 takes 10+50/1.2 = 10+ 41 2/3=51 2/3

run 20 swim 40sqr2 then 6 2/3 +47.1 = 53.8 sec

run 50 swim sqr1700 then 16 2/3 +34.359= 51.025 sec < 51.271 though