Math Application Help Please?? :-)?

Question

This is the only problem I can't figure how I got wrong on my last test.

1.) A lakefront runs east-west. A man in a rowboat is 5 miles due north of point A on the shore. He wishes to get to B, 5 miles due east of A, in the least time. He is able to row 3 miles per hour and walk 5 hour. What is the minimum time for the trip in minutes?

Answer 1

Square root of 50 / 3 * 60 = minutes

Answer 2

Let
x = distance west of point B where boat landed

w = distance traveled in water
x = distance traveled over land

By the Pythagorean Theorem

w² = 5² + (5 - x)² = 25 + 25 - 10x + x²
w² = x² - 10x + 50
w = √(x² - 10x + 50)

Time traveled.

t = w/3 + x/5 = √(x² - 10x + 50) / 3 + x/5

Take the derivative and set equal to zero to find the critical points.

dt/dx = (1/2)(2x - 10)/[3√(x² - 10x + 50)] + 1/5 = 0
(x - 5)/[3√(x² - 10x + 50)] = -1/5
5(x - 5) = -3√(x² - 10x + 50)

Square both sides.

25(x - 5)² = 9(x² - 10x + 50)
25(x² - 10 x + 25) = 9(x² - 10x + 50)
25x² - 250 x + 625 = 9x² - 90x + 450
16x² - 160 x + 175 = 0
(4x - 5)(4x - 35) = 0

x = 5/4, 35/4

But x ≤ 5 so the only solution is:
x = 5/4

Plug in and solve for time.

t = w/3 + x/5 = √(x² - 10x + 50) / 3 + x/5
t = √[(5/4)² - 10(5/4) + 50] / 3 + (5/4)/5
t = √[25/16 - 25/2 + 50] / 3 + 1/5
t = √[(25 - 8*25 + 16*50)/16] / 3 + 1/5
t = √[625/16] / 3 + 1/5
t = (25/4)/3 + 1/5 = 25/12 + 1/5 = (125 + 12)/60
t = 137/60 hours = 137 minutes

The minimum time for the trip is 137 minutes.

Answer 3

The answer is 140 minutes.

The man has the option to land at any point C between A and B, then walk the distance from C to B.

Let W represent the distance from C to B, which he will cover by walking at 5 miles per hour.

Let R represent the distance from his initial position to C, which he will cover by rowing at 3 miles per hour.

Now R is the hypotenuse of a right triangle whose 2 legs are 5 and (5-W)

By pythagorean theorem,
R = sqrt[5^2 + (5-W)^2] = sqrt ( W^2 -10W + 50 )

The total time = time spent rowing + time spent walking
t = R/3 + W/5

t = [ sqrt(W^2 -10W + 50) ] / 3 + W/5

take the derivative of t with respect to W, and set it = 0, to get the value of W that yields the least amount of time:

dt/dW = (1/3)*(1/2)*[ (W^2 -10W + 50)^(-1/2) ]*(2W-10) + 1/5
0 = (1/6)*[ (W^2 - 10W + 50)^(-1/2) ]*(2W-10) + 1/5

After expanding, collecting like terms, we get
64W^2 - 640W + 700 = 0
16W^2 -160W + 175 = 0

Using quadratic formula, we get 2 values of W:
W = 8.75 miles or W = 1.25 miles

But W cannot be longer than 5, so we take W = 1.25 miles

Solving for R, we get R = 6.25

Therefore the minimum time for the trip occurs when W = 1.25 and R = 6.25

t = R/3 + W/5
t = (6.25/3 + 1.25/5) hrs
t = (6.25/3 + 1.25/5) hrs * 60 minutes/hr
t = 140 minutes

Answer 4

= ([√{5 mi^2 + 5 mi^2} / 3 mph] + 5 hr) * 60 min
= ([√{25 mi + 25 mi} / 3 mph] + 5 hr) * 60 min
= ([√50 mi / 3 mph] + 5 hr) * 60 min
= ([7.071 / 3 mph] + 5 hr) * 60 min
= (2.357 hr + 5 hr) * 60 min
= 7.357 hr * 60 min
= 441.42 min

Answer: 441 minutes

He rows 2.357 hours or 141 minutes and walks 5 hours or 300 minutes for a total of 7.357 hours or 441 minutes.

I hope this mends a broken heart.