Question

The shortest distance across the water from an island dock (point D) to a straight shoreline is 9 miles. Let the closest point on the shoreline be A. An amphibious vehicle makes a trip from a depot at a point B on the shoreline that is 20 miles away from A. Suppose that it travels 20 – x miles toward A along the shoreline (where 0 ≤ x ≤ 20) to a point C, then strikes out across the water toward the dock D. The vehicle can travel 30 miles per hour on land and 12 miles per hour across water.

a.Draw a diagram and mark on it the distances AD, AB, BC, AC, and CD, expressing distances that depend on x in terms of x.

b.Show that the time required to travel from the depot to the dock is given by T(x)=2(20-x)+5√(x^2+81) (in minutes).

c.Determine the distance the vehicle should travel on land in order to minimize the time for the trip. What is the minimum time?

Answer 1

18tax N shore A4 . с до-х в AD=9 Distance traveled from B to c is ao-x so BC - 20-X AB= 20 also AB=ACTOBC 20= AC & do-X AC = 20-20+X =X CD is the hypothuse of right angle formed by ao AC=x and AD=9. so CD = 592+x2 = √x281 triangle

b) we knowe Time - Distance speed. find time for both past of Journey then total time is sum of this two time shore line Distance = 20 x Speed = 30 mph so time = 20-2 azo In hour Water Distance = x2+81 Speed =12 So time = V2781 in hour - 12 Total time 700) = 20-2 + x 2781 in houy 12 take LCM of denominator and make bom into some donominator. LEM (30,12) =60. so

Multiply numerator and denominator of First term by 2 and second term by s 12 so 2 20-2) + 51x2781 60 60 TO) = - 2 (20-8) + 5 500781 (in hour) 60 This is time in hour: to make it in to time in minute multiply by 60. T(x) = 2 (20-x) + 5 x ² +81 o so Tec) in minute is T(x) = 2 (20-x) + 5 x ²78)

c) Now" Lets find minimum of T60) ". Minimum of TO) occur when T1(x) is zero. Thorax 2 (20-x)+5 12281) poc a (0-1) + 5032781 co (40)-da ax + 5 de 122781 We know constant = 0, X=. so al lade D22781 = y 24 l Th@C) = 0–2 + Now eausse TIGC) equel to zero - 27 500 - 0 2+8)

5x = 2√x2+81 squire both side. assant (5x)² = (2x ?+81) ² 25 x ² = 4 (00²781) 25x2 = 4x² + 324 2500² - 4x² = 324 21xc2 = 324 x2 = 324 21 x= √324 = 3.92792 'Y 3.928 distance travelled on shore = 20-x . = 20 - 3.928 - 16.072 ] So he should travel 16.019 miles on land. (note it is approximately a 16 miles). to find minnum time find T (3428) T63-928 = 2 (20-3.928) + 5 /3.928781 = 81.243. Minimum time = 81.243 minute

to find last part you can also do by graphing. Graph T(x) and find least point . Least point gives x and minimum time .