Question

Sally is standing at one corner of a snowy rectangular field that measures 200 ft by 200 ft. She wishes toreach a warm cabin located at the diagonally opposite corner of the field. Suppose Sally trudges throughthe snow at 3 ft/sec to reach pointPat the other side of the field, and then walks the remaining distance along a paved road at 5 ft/sec to reach the cabin. What is the optimal point that will allow Sally toreach the cabin as quickly as possible

Answer 1

Complete Question

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Answer:

The optimal point(i.e the position of  P) that will allow Sally to reach the cabin as quickly as possible is  $d = 50 \ ft$ from the cabin.

Step-by-step explanation:

From the question we are told that

   The speed from the point Sally is standing to the point P is  $u = 3 \ ft /s$

   The speed from the point P to the cabin is $v = 5 \ ft /s$

   

Let denote the distance from the bottom left corner to the  P  be  y  ft

 Generally according to Pythagoras theorem the distance from Sally's first position to point P is mathematically represented as

          $s = \sqrt{y^2 + 200^2}$

Generally the distance from that point P to the cabin is mathematically represented as

       $d = 200 -y$

Generally the time it takes Sally to cover that distance s is mathematically represented as

       $t_1 = \frac{s}{u}$

=>     $t_1 = \frac{ \sqrt{y^2 + 200^2}}{3}$

Generally the time it takes Sally to cover that distance d is mathematically represented as

       $t_2 = \frac{d}{v}$

=>     $t_2 = \frac{ 200 - y }{5}$

So the total time taken is  

        $t= \frac{ \sqrt{y^2 + 200^2}}{3} + \frac{ 200 - y }{5}$

Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0

So  

     $\frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} } - \frac{1}{5} =0$      

=> $\frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}$

=> $25y^2 = 9y^2 + 9 * 200^2$        

=> $16y^2 = 360000$          

=> $y = 150$

So the distance from point P to the cabin is  

           $d = 200 -150$

=>       $d = 50 \ ft$

So  the optimal point(i.e the position of  P) that will allow Sally to reach the cabin as quickly as possible is  $d = 50 \ ft$ from the cabin