Respuesta :
Answer:
The visitor should run approximately 14.96 mile to minimize the time it takes to reach the island
Step-by-step explanation:
From the question, we have;
The distance of the island from the shoreline = 1 mile
The distance the person is staying from the point on the shoreline = 15 mile
The rate at which the visitor runs = 6 mph
The rate at which the visitor swims = 2.5 mph
Let 'x' represent the distance the person runs, we have;
The distance to swim = [tex]\sqrt{(15-x)^2+1^2}[/tex]
The total time, 't', is given as follows;
[tex]t = \dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}[/tex]
The minimum value of 't' is found by differentiating with an online tool, as follows;
[tex]\dfrac{dt}{dx} = \dfrac{d\left(\dfrac{x}{6} +\dfrac{\sqrt{(15-x)^2+1^2}}{2.5}\right)}{dx} = \dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} }[/tex]
At the maximum/minimum point, we have;
[tex]\dfrac{1}{6} -\dfrac{6 - 0.4\cdot x}{\sqrt{x^2-30\cdot x +226} } = 0[/tex]
Simplifying, with a graphing calculator, we get;
-4.72·x² + 142·x - 1,070 = 0
From which we also get x ≈ 15.04 and x ≈ 0.64956
x ≈ 15.04 mile
Therefore, given that 15.04 mi is 0.04 mi after the point, the distance he should run = 15 mi - 0.04 mi ≈ 14.96 mi
[tex]t = \dfrac{14.96}{6} +\dfrac{\sqrt{(15-14.96)^2+1^2}}{2.5} \approx 2..89[/tex]
Therefore, the distance to run, x ≈ 14.96 mile
