# How to do question 10?

May 20, 2018

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May 20, 2018

C, more specifically $3$ units below B, to ensure the shortest time taken to reach the injured.

#### Explanation:

Let the time taken to reach the injured be $t$ and the distance they land the boat from B be $x$, and we can form an equation that follows,

$t = \frac{\sqrt{{4}^{2} + {x}^{2}}}{6} + \frac{10 - x}{10}$

where $\sqrt{{4}^{2} + {x}^{2}}$ is the distance travelled by boat, and $10 - x$ is the distance travelled on foot.

Simplify,

$t = \frac{1}{6} {\left(16 + {x}^{2}\right)}^{\frac{1}{2}} - \frac{1}{10} x + 1$

Differentiate,

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{6} \cdot \frac{1}{2} \cdot 2 x \cdot {\left(16 + {x}^{2}\right)}^{- \frac{1}{2}} - \frac{1}{10}$

Simplify,

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{2 x}{12 \sqrt{16 + {x}^{2}}} - \frac{1}{10}$

Let $\frac{\mathrm{dt}}{\mathrm{dx}} = 0$

$\frac{2 x}{12 \sqrt{16 + {x}^{2}}} - \frac{1}{10} = 0$

Add $\frac{1}{10}$ to both sides,

$\frac{2 x}{12 \sqrt{16 + {x}^{2}}} = \frac{1}{10}$

Cross multiply,

$20 x = 12 \sqrt{16 + {x}^{2}}$

Square both sides,

$400 {x}^{2} = 144 \left(16 + {x}^{2}\right)$

Expand,

$400 {x}^{2} = 2304 + 144 {x}^{2}$

Subtract $144 {x}^{2}$ from both sides,

$256 {x}^{2} = 2304$

Divide both sides by $256$

${x}^{2} = 9$

Square root both sides,

$x = 3$ ( reject $- 3$ as distance$> 0$ )

Hence, they should land $3$ units from B, which is C.