Question

Mauritius island is 2 km due south of its closest point along a straight coastline. Gibril is staying at a private room on the shore that is 6 km east of that point. Gibril is planning to go from the room to the island. Suppose Gibril runs at a rate of 8 kmph and swims at a rate of 3 kmph. How far should Gibril run before swimming to minimize the time it takes to reach the island?

Answer 1

Answer:

5.19 Km

Step-by-step explanation:

According To the Question,

Suppose he Runs Xkm before swimming to minimize the time it takes to reach the island .

So, he should swim $\sqrt{2^{2}+(6-X)^{2} } Km$ .

Now The Total Time is t = $\frac{X}{8} + \frac{\sqrt{2^{2}+(6-X)^{2} } }{3}$ .

t = $\frac{X}{8} + \frac{\sqrt{X^{2}-12X+40 } }{3}$

$\frac{dt}{dX}=1/8 + 1/3 * 1/2 * \frac{2X-12}{\sqrt{X^{2}-12X+40 } }$

$\frac{dt}{dX}=1/8 + \frac{1}{3} * \frac{X-6}{\sqrt{X^{2}-12X+40 } }$

Now Put $\frac{dt}{dX} = 0$ , we get

$1/8 + \frac{1}{3} * \frac{X-6}{\sqrt{X^{2}-12X+40 } } = 0$

$\frac{1}{3} * \frac{X-6}{\sqrt{X^{2}-12X+40 } } = - \frac{1}{8}$

8(X-6)= -3 × ${\sqrt{X^{2}-12X+40 }$

square on both side,We get

64(X²-12X+36) = 9X² - 108X +360

55X² -660X + 1944 = 0

Apply Shri Dharacharya formula, To Find The Value of 'X'

X =[ - (-660) ±$\sqrt{(-660)^{2}-4*55*1944 }$  ] ÷ 2×55

X = (660±√7920) / 110

X= (660 - 89 ) / 110  ⇒ (Neglect '+' because it give 749/110 = 6.8 km Which is not possible )

X=571/110 ⇔ 5.19km he Should Run before swimming to minimize the time .

(For Diagram,Please Find In Attachment)