Question

A boat on the ocean is 2 mi from the nearest point on a straight? shoreline; that point is 15 mi from a restaurant on the shore. A woman plans to row the boat straight to a point on the shore and then walk along the shore to the restaurant. Complete parts?(a) and? (b) below.

a. If she walks at 3? mi/hr and rows at 2? mi/hr, at which point on the shore should she land to minimize the total travel? time?

Let x be the distance between the nearest point on shore and the point she lands on shore. If T is the time it takes her to get to the? restaurant, what is the objective ?function?

b. If she walks at 3? mi/hr, what is the minimum speed at which she must row so that the quickest way to the restaurant is to row directly? (with no? walking)?

Answer 1

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

Answer 2

A) x is = 0,70 miles

T = 1/2* (√4 + x²) +( 15 - x )/3 Objective Function to minimize

B) V(min) is = 2.59 m/hr

Calculation of Travel and Time

Let's assume the boat is in Point A, she lands in point B, and also R is the restaurant.

Then, Let's call x distance between the point O in which the perpendicular line from the boat gets to the shoreline, and also the point where she land.

Then, We know distance is

That is: d = v*t ⇒ t = d/v

Then, She rows at 2 miles/hr and walk at 3 miles /hr

Now, According to that she takes

c (hypotenuse) of right triangle AOB/ 2 rowing

and 15 - x /3 walking

Total time is:

Then T = c/2 + ( 15-x )/3

Then c =√(2)² + x² ⇒ T = [√(2)² + x² ] /2 + ( 15-x )/3

After that T = 1/2* (√4 + x²) +( 15 - x )/3 (1)

And also, that is the Objective function to minimize

A. When we Taking derivatives on both sides of the equation we get

Now, 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

Then, 9x² = 4 + x² ⇒ 8x² = 4 x² = 1/2 x = 0,70 miles

B. Now, If we plug this value in the Objective function we will get the minimum time

After that, 1/2* (√4 + x²) +( 15 - x )/3 ⇒ [√ 4 + 0,5] /2 + 14,3/3 = T (min)

Then T(min) = 1.06 + 4.77 = 5.83 hr

When the Distance L (hypotenuse of right triangle AOR)

Then L = √(2)² + (15)² L = 15,13 miles

And also that distance have to be traveled at least in 5.83 hr rowing

Therefore, as v = d/t V(min) = 15.13/ 5.83 ⇒ V(min) = 2.59 m/hr

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