Question

Required information Problem 11.005 - Altitude as a function of time - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS NOTE: This is a multi-part question. Once an answer is subted, you will be unable to return to this part A group of hikers uses a GPS while doing a 40-mile trek in Colorado. A curve fit to the data shows that their altitude can be approximated by the function y(t) -0.12 -6.75 + 135 - 11206 + 3200€ + 9060 where y and are expressed in feet and hours, respectively Problem 11.005.a - Finding maximum altitude During the 18-hour hike, determine the maximum altitude that the hikers reach (You must provide an answer before moving on to the next part.) The maximum altitude is it < Prev 3 4 5 of 95 !!! Next >

Answer 1

Solution : Criven! y(t) = 0.12 95 -6.75 44+ 13573 - 112072 +32004 +9060 During 18 hour hike - Maximum value of y(t) & find out. Using the concept of minima and maxima, first of all we have to find critical points. y +)20 Yct = 0.627 - 27ts & 6051²-22404 +3200=6 24 After solving Biquadratic.. t = 2.15 8.51, 15.59, 18.67 in hours To determine which is minima and which is maxima. y (+) > 01 3 point of Minima? lat point 9" (+) colodint & Point of maxima? Moxima . Minima points 200.6 0221 points 2.15 h reordnessa 8.51 h 15.59 h 18.67 X h. 3 out of Raupe? of hikers? Not at = 2.15h Y(+) = 012(2.15)5 - 6.75 (2.1599 + (135) (2.15) - 1120 (2015)2 + 3200 (2.15) +9060 y (2.15) = 11965.98 ft

At 42 15:59h. y (15:59) = 0.12 (15:59) -6.75 (15.59) + ( 135 (15.5913 – 1420 (15:59)² + 3200 (15:59) + 90601 de y ( 15.59) - 10254.66 ft. But value we have to find of alltitude lie the maximum Global maximas. 19(2015) =11965.98 ft 11965.98 ft I is maximum altitude X during the journey of 18h [11965-98f7 Required Aus. я
Solved using Maxima minima concept that is global Maxima....