Can anyone figure out this optimization problem and show the work?

Devin asked:

A lighthouse is at point P, 3 miles offshore, from the nearest point X of a straight beach. A store is located 5 miles down the beach from x The lighthouse keeper acan row 3 miles/hour and walk 4 miles /hour. How far along the beach from X should he land in order to get to the store in the shortest possible time?

Posted by admin on Tuesday, June 30, 2009 at 9:57 pm.

Filed under Mathematics, Lighthouse Keeper, Offshore, Optimization Problem.

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3 Comments on “Can anyone figure out this optimization problem and show the work?”

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Math Help Guy says:

Let d be the distance between point X and where the lighthouse keeper lands on shore.

His distance traveled by boat is sqrt(3^2 + d^2).
His distance traveled by foot is 5 - d.

His time by boat is sqrt(9 + d^2)/3
His time by foot is (5 - d)/4

We need the sum of these times to be a minimum.

time, written as a function of distance, d, is:

t(d) = (1/3)sqrt(9 + d^2) + (1/4)(5 - d)

Let’s find t’(d), then solve t’(d) = 0 for d and we will have your answer…

t’(d) = (1/3)d(9+d^2)^(-1/2) - 1/4

(1/3)d(9+d^2)^(-1/2) - 1/4 = 0

(1/3)d(9+d^2)^(-1/2) = 1/4

d / sqrt(9 + d^2) = 3/4

Cross multiply:

4d = 3sqrt(9 + d^2)

Square both sides:

16d^2 = 9(9+d^2)

16d^2 = 81 + 9d^2

7d^2 = 81

d^2 = 81/7

d = sqrt(81/7) = 9/sqrt(7) = (9/7)sqrt(7) = 3.4

t’ = 0 when d = (9/7)sqrt(7)

Answer: d = 3.4 miles from point X.

July 2nd, 2009 at 3:07 pm
miryana kina says:

let x be the distance he’d walk and t(x) be the function of time needed.

t(x) = x/4 + √[3^2 + (5 - x)^2] /3
= x/4 + √(x^2 - 10x + 34) /3

t’(x) = 1/4 + (2x - 10) / 2*3*√(x^2 - 10x + 34)
= 1/4 + (x - 5)/3√(x^2 - 10x + 34) = 0
9(x^2 - 10x + 34) = 16(x^2 - 10x + 25)
9x^2 - 90x + 306 = 16x^2 - 160x + 400
7x^2 - 70x + 94 = 0
7(x - 5)^2 = 81
(x - 5)^2 = 81/7
x = 5 +- 9/√7
x = distance, so 0

July 2nd, 2009 at 6:57 pm
cidyah says:

Let y be the distance from the shore to point x. Then 5-y is the distance from x to the store.
time = distance / speed
rowing: distance = sqrt(y^2+9) : speed = 3 miles/hr : time = sqrt(y^2+9) / 3
Walking distance =5-y : speed = 4 miles/hr : time : (5-y)/4
Total time t= sqrt(y^2+9) / 3 + (5-y) / 4
dt/dy = 2y/2(3)sqrt(y^2+9) -1/4 =0
y / 3sqrt(y^2+9) - 1/4 =0
4y - 3sqrt(y^2+9) = 0
16y^2= 9(y^2+9)
7y^2=81
y^2=81/7
y=9/sqrt(7)=3.4 miles

July 4th, 2009 at 5:13 am