A lighthouse is built on an exposed reef, 4.5 miles off-shore. The shoreline is perfectly straight, and a town is located 6 miles downshore from the point on the shoreline nearest the lighthouse.
The lighthouse keeper needs to go from the lighthouse to the town to get fresh supplies. He can row a boat at 1.8 miles per hour, and run at 3.4 miles per hour.
How far downshore towards the town should he land, if he wants to get to the town as quickly as possible?

3.94 hours

Step-by-step explanation:

From the question, the following parameters are given:

Distance offshore = 4.5 miles

Distance downshore = 6 miles

Speed when running = 3.4 mph

Speed with boat = 1.8 mph

distance of boat rowing = sqrt(4.5^2 + x^2)

Where

Speed = distance/time

Time = distance/speed

Time = distance of boat rowing/1.8

distance of running = 6 - x

Time = (6-x)/3.4

total travel time

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

dt/dx = x/(1.8×sqrt(4.5^2 + x^2) - 1/3.4

d^2t/dx^2 = +ve at any x

x is at its minimum when dt/dx=0

x/(1.8×sqrt(4.5^2 + x^2) = 1/3.4

x = 6/sqrt(253) × 4.5 = 1.697 miles

Substitute x in

t = sqrt(4.5^2 + x^2)/1.8 + (6 - x)/3.4

We obtaine

t = sqrt(4.5^2 + 1.697^2)/1.8 + (6 - 1.69)/3.4

t = 2.672+ 1.266 = 3.94 hours

what is the starting amount? the question is not clear sorry.

step-by-step explanation:

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse