Answer:
a) x = 0,70 miles
T = 1/2* (√4 + x²) +( 15 - x )/3 Objective Function to minimize
b) V(min) = 2.59 m/hr
Step-by-step explanation:
Let assume the boat is in Point A, she lands in point B, and R is the restaurant
Let call x distance between the point O in which perpendicular line from the boat get to shoreline, and the point were she land.
We know distance is
d = v*t ⇒ t = d/v
She rows at 2 miles/hr and walk at 3 miles /hr
According to that she takes
c (hypotenuse) of right triangle AOB/ 2 rowing
and 15 - x /3 walking
Total time is:
T = c/2 + ( 15-x )/3 c =√(2)² + x² ⇒ T = [√(2)² + x² ] /2 + ( 15-x )/3
T = 1/2* (√4 + x²) +( 15 - x )/3 (1)
And that is the Objective function to minimize
a) Taking derivatives on both sides of the equation we get
T´(t) = x /( √4 + x²) - 1/3 ⇒ T´(t) = 0 x /( √4 + x²) - 1/3 = 0
3*x - √( √4 + x²) = 0 ⇒ 3*x = √( √4 + x²)
Squaring both sides
9x² = 4 + x² ⇒ 8x² = 4 x² = 1/2 x = 0,70 miles
If we plug this value in the Objective function we will get the minimum time
1/2* (√4 + x²) +( 15 - x )/3 ⇒ [√ 4 + 0,5] /2 + 14,3/3 = T (min)
T(min) = 1.06 + 4.77 = 5.83 hr
Distance L (hypotenuse of right triangle AOR)
L = √(2)² + (15)² L = 15,13 miles
And that distance have to be traveled at least in 5.83 hr rowing
Then as v = d/t V(min) = 15.13/ 5.83 ⇒ V(min) = 2.59 m/hr
