Answer to Question #98488 in Calculus for abe

Question

Maya is 2 km offshore in a boat and wishes to reach a coastal village which is 6 km down a straight
shoreline from the point on the shore nearest to the boat. She can row at 2 km/hr and run at 5
km/hr. Where should she land her boat to reach the village in the least amount of time?

Answer 1

Finding the Unknown value in the word problem

We are going to find where, Maya land her boat to reach the village in the least amount of time

Solution:

Let C = 2 km and b = 6 km

Speed at running = 5 kmph

speed to walk = 2kmph

By Pythagorean fomrula

$$l^2 = c^2 + x^2$$

$$l = \sqrt {4+x^2}$$

Time required to cross the river

$$Time = \frac {Distance } { speed} = \frac {l} {2} = \frac {\sqrt {4 +x^2}} {2}$$

Maya should run the distance =

$$b - x = 6 - x$$

Time for run =

$$\frac {Distance} {speed} = \frac {6-x}{5}$$

Total time =

$$t = \frac {1}{2} \times \sqrt {x^2+4} + \frac {6-x} {5}$$

$$t = \frac {1}{2} \times {(x^2+4)}^{\frac {1}{2}} + \frac {6-x} {5}$$

Differentiate with respect to x,

$$\frac {dt}{dx} = \frac{1}{2} \times \frac {1}{2} \times {(x^2+4)}^{\frac {1}{2}-1} (2x) - \frac {1}{5}$$

$$\frac {dt}{dx} = \frac {x}{2} \times {(x^2+4)}^{\frac {1}{2}-1} - \frac {1}{5}$$

$$\frac {dt}{dx} = \frac {x}{2} \times {(x^2+4)}^{\frac {-1}{2}} - \frac {1}{5}$$

$$\frac {dt}{dx} = \frac {x}{2} \times \frac {1} {\sqrt {(x^2+4)}} - \frac {1}{5}$$

Now set the derivative to zero, to find the minimum value of x

$$\frac {x}{2} \times \frac {1} {\sqrt {(x^2+4)}} - \frac {1}{5} =0$$

$$\frac {x} {\sqrt {(x^2+4)}} = \frac {2}{5}$$

Squaring on both the sides,

$$\frac {x^2} {{(x^2+4)}} = \frac {4}{25}$$

$$25x^2 = 4 (x^2 + 4)$$

$$25x^2 = 4x^2 + 16$$ $$21x^2 = 16$$

$$x^2 = \frac {16} {21}$$

$$x = \frac {4} {\sqrt 21}$$

Answer:

At $x = \frac {4} {\sqrt 21}$ km down the shore she land her boat to reach the village.