)an amphibious vehicle requires 5 minutes to travel each mile over land, but only 4 minutes to travel each mile in the water. the vehicle is currently located on water, in a canal that runs in the east-west direction. its destination point is on land, 3 miles due east and 2 miles due north of its current position. what route will minimize the time required for the vehicle to reach its destination? (you should assume that the time required for the vehicle to transition from water to land is negligible, and that the canal’s width is also negligible.) justify your answer completely.

x = 0.63 miles

y = 3.10 miles

Step-by-step explanation: See Annex

Vehicle capacities.

In land 1 mile in 5 minutes    then   1/5 mile in 1 minute

In water 1 mile in 4 minutes   then  1/4 mile in 1 minute

Let call x the number of miles in water and y the number of miles in land

Then the route will be x (miles in water ) and y miles in land

From Annex

L = 1/4*x + 1/5*y

From Figure (Annex) y = √( 3 - x )² + ( 2)² Pithagoream theorem

y = √ 9 + x² -6*x + 4 ⇒ y = √ x² - 6*x + 13  

Then

L(x) = 1/4x + 1/5√ x² - 6*x + 13  

Taking derivatives on both sides of the equation:

L´(x) = 1/4 + 1/5* 1/2 (2*x -6) / √ x² - 6*x + 13  

L´(x) = 0 ⇒ 1/4 + 1/5* 1/2 (2*x -6) / √ x² - 6*x + 13 = 0

[1/4* √( x² - 6*x + 13) + 1/5* (x -3) ] /√ x² - 6*x + 13 = 0  

1/4* √( x² - 6*x + 13) + 1/5*x - 3/15 = 0      ⇒ 1/4* √( x² - 6*x + 13)  = 1/5  ( x -3)

1/4* √( x² - 6*x + 13) = 1/5*x  - 1/5      

Squaring both sides

1/16* (x² - 6*x + 13) = 1/25 (x - 3 )²   ⇒   x²/16  - 6/16 *x + 13/16 = 1/25 ( x² + 9 - 6*x

25 (x² - 6*x + 13)  = 16 ( x² + 9 -6*x)

25*x² - 150*x +  325  =  16x² + 144 - 96*x

9*x² - 294*x + 181 =  0

x₁,₂  =  294 ± √ (86436 -  6516) / 18

x₁   =  294 +  282,7 / 18      we dismiss that solution si not feasible with problem statemente

x₂ = 294 - 282,7 / 18

x₂ = 11.3/ 18

x₂ =  0.63 miles      and  y =  √ x² - 6*x + 13     ⇒    y  = 3.10 miles

there is no question content here, but you for the free points! : d