Complete Question
The complete question is shown on the first uploaded image
Answer:
The optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is [tex]d = 50 \ ft[/tex] 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 [tex]u = 3 \ ft /s[/tex]
The speed from the point P to the cabin is [tex]v = 5 \ ft /s[/tex]
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
[tex]s = \sqrt{y^2 + 200^2}[/tex]
Generally the distance from that point P to the cabin is mathematically represented as
[tex]d = 200 -y[/tex]
Generally the time it takes Sally to cover that distance s is mathematically represented as
[tex]t_1 = \frac{s}{u}[/tex]
=> [tex]t_1 = \frac{ \sqrt{y^2 + 200^2}}{3}[/tex]
Generally the time it takes Sally to cover that distance d is mathematically represented as
[tex]t_2 = \frac{d}{v}[/tex]
=> [tex]t_2 = \frac{ 200 - y }{5}[/tex]
So the total time taken is
[tex]t= \frac{ \sqrt{y^2 + 200^2}}{3} + \frac{ 200 - y }{5}[/tex]
Generally the minimum time taken with respect to the distance is mathematically evaluated by differentiating and equating the derivative to 0
So
[tex]\frac{dt}{dy} =\frac{1}{3} \frac{y}{ \sqrt{y^2 + 200^2} } - \frac{1}{5} =0[/tex]
=> [tex]\frac{dt}{dy} = \frac{y^2 }{y^2 + 200 ^2} =\frac{9}{25}[/tex]
=> [tex]25y^2 = 9y^2 + 9 * 200^2[/tex]
=> [tex]16y^2 = 360000[/tex]
=> [tex]y = 150[/tex]
So the distance from point P to the cabin is
[tex]d = 200 -150[/tex]
=> [tex]d = 50 \ ft[/tex]
So the optimal point(i.e the position of P) that will allow Sally to reach the cabin as quickly as possible is [tex]d = 50 \ ft[/tex] from the cabin
