Answer:
This is a classic optimization problem in calculus. Let's denote the dog's swimming speed as
�
s and the dog's running speed as
3
�
3s. Also, let
�
x be the distance along the shoreline where the dog exits the water and starts running.
The total time
�
T for the dog to get to the town is the sum of the time spent swimming and the time spent running:
�
=
10
�
+
40
−
�
3
�
T=
s
10
+
3s
40−x
To minimize
�
T, we need to find the value of
�
x that minimizes
�
T.
�
=
10
�
+
40
−
�
3
�
T=
s
10
+
3s
40−x
Now, we can take the derivative of
�
T with respect to
�
x and set it equal to zero to find the critical point:
�
�
�
�
=
0
dx
dT
=0
�
�
�
(
10
�
+
40
−
�
3
�
)
=
0
dx
d
(
s
10
+
3s
40−x
)=0
Solving for
�
x will give us the optimal point along the shoreline where the dog should start running to minimize the total time.