question 10 - 1 point mia is staying in a cottage along a beautiful and straight shoreline. a point q on the…

question 10 - 1 point mia is staying in a cottage along a beautiful and straight shoreline. a point q on the shoreline is located 4 kilometers east of the cottage, and an island is located 2 kilometers north of q (see image below). mia plans to travel from the cottage to the island by some combination of walking and swimming. she can start to swim at any point p between the cottage and the point q. if she walks at a rate of 6 km/hr and swims at a rate of 4 km/hr, what is the minimum possible time it will take mia to reach the island? enter an exact answer or round to the nearest hundredth of an hour.

question 10 - 1 point mia is staying in a cottage along a beautiful and straight shoreline. a point q on the shoreline is located 4 kilometers east of the cottage, and an island is located 2 kilometers north of q (see image below). mia plans to travel from the cottage to the island by some combination of walking and swimming. she can start to swim at any point p between the cottage and the point q. if she walks at a rate of 6 km/hr and swims at a rate of 4 km/hr, what is the minimum possible time it will take mia to reach the island? enter an exact answer or round to the nearest hundredth of an hour.

Answer

Explanation:

Step1: Define variables

Let $x$ be the distance Mia walks along the shoreline. Then the distance she swims is $\sqrt{(4 - x)^2+2^2}$ according to the Pythagorean theorem.

Step2: Set up time - function

The time $T$ taken is the sum of the walking time and the swimming time. The walking time $t_1=\frac{x}{6}$ and the swimming time $t_2=\frac{\sqrt{(4 - x)^2 + 4}}{4}$. So $T(x)=\frac{x}{6}+\frac{\sqrt{(4 - x)^2+4}}{4}$.

Step3: Differentiate the time - function

First, expand $(4 - x)^2=16-8x + x^2$. Then $T(x)=\frac{x}{6}+\frac{\sqrt{x^{2}-8x + 20}}{4}$. Using the chain - rule, $T^\prime(x)=\frac{1}{6}+\frac{2x - 8}{8\sqrt{x^{2}-8x + 20}}$.

Step4: Set the derivative equal to zero

$\frac{1}{6}+\frac{2x - 8}{8\sqrt{x^{2}-8x + 20}}=0$. Cross - multiply to get $8\sqrt{x^{2}-8x + 20}+6(2x - 8)=0$. Then $8\sqrt{x^{2}-8x + 20}=-12x + 48$. Square both sides (note: we need to check for extraneous solutions later): $64(x^{2}-8x + 20)=144x^{2}-1152x + 2304$. Expand: $64x^{2}-512x+1280 = 144x^{2}-1152x + 2304$. Rearrange to get $80x^{2}-640x + 1024 = 0$. Divide by 16: $5x^{2}-40x + 64 = 0$. Using the quadratic formula $x=\frac{40\pm\sqrt{1600 - 1280}}{10}=\frac{40\pm\sqrt{320}}{10}=\frac{40\pm8\sqrt{5}}{10}=4\pm\frac{4\sqrt{5}}{5}$. We take $x = 4-\frac{4\sqrt{5}}{5}$ since $x\in[0,4]$.

Step5: Calculate the minimum time

Substitute $x = 4-\frac{4\sqrt{5}}{5}$ into $T(x)$. $T\left(4-\frac{4\sqrt{5}}{5}\right)=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{\sqrt{\left(4-(4 - \frac{4\sqrt{5}}{5})\right)^2+4}}{4}$ $=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{\sqrt{\frac{16}{5}+4}}{4}=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{\sqrt{\frac{16 + 20}{5}}}{4}=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{\sqrt{\frac{36}{5}}}{4}$ $=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{\frac{6}{\sqrt{5}}}{4}=\frac{4-\frac{4\sqrt{5}}{5}}{6}+\frac{3}{2\sqrt{5}}$ $=\frac{20 - 4\sqrt{5}}{30}+\frac{9\sqrt{5}}{30}=\frac{20 - 4\sqrt{5}+9\sqrt{5}}{30}=\frac{20 + 5\sqrt{5}}{30}=\frac{4+\sqrt{5}}{6}\approx1.19$ hours.

Answer:

$1.19$ hours