Hello, afan17!

I'm new to this forum . . . Welcome aboard!

I'm actually tutoring a refugee in a volunteer program . . . Good for you!

A resort has a rectangular swimming pool ABCD with AB = 75m, AD = 30m.

P is a point somewhere between D and C.

A boy can swim at 1 m/s and run at 1 2/3 m/s.

He starts at A, swims to point P, and runs from P to C.

He takes 2 seconds to pull himself out of the pool.

a) Let DP = x meters and the total time be T seconds.

Show that: .$\displaystyle T\:=\:\sqrt{x^2+900}+ \frac{3}{5}(75-x)+2$ Code:

A 75 B
* - - - - - - - - - - - - - - *
| * |
| * |
30 | * |
| * |
| * |
* - - - - - - - - * - - - - - *
D x P 75-x C

He will swim the diagonal distance $\displaystyle AP.$

This is the hypotenuse of right triangle $\displaystyle ADP.$

. . Hence:. . $\displaystyle AP \:=\:\sqrt{x^2 + 30^2}$

At 1 m/sec, this takes him: .$\displaystyle \frac{\sqrt{x^2+900}}{1} \:= \:{\color{blue}\sqrt{x^2+900}}$ seconds.

He runs the distance $\displaystyle PC \:=\:75-x\,\text{ at }\,\frac{5}{3}$ m/sec.

This takes him: .$\displaystyle \frac{75-x}{\frac{5}{3}} \:=\:{\color{blue}\frac{3}{5}(75-x)}$ seconds.

He also used 2 seconds to leave the pool.

His total time is: .$\displaystyle \boxed{T \;=\;\sqrt{x^2+900} + \frac{3}{5}(75-x) + 2}$ seconds.

b) Find $\displaystyle \frac{dT}{dx}$ We have: .$\displaystyle T \;=\;\left(x^2+900\right)^{\frac{1}{2}} - \frac{3}{5}x + 47$

Then: .$\displaystyle \frac{dT}{dx} \;=\;\frac{1}{2}\left(x^2+900\right)^{-\frac{1}{2}}(2x) - \frac{3}{5} \;=\;\boxed{\frac{x}{\sqrt{x^2+900}} - \frac{3}{5}}$

c) (i) Find the value of $\displaystyle x$ for which $\displaystyle T$ is a minimum.

(ii) Find the minimum time. (i) Solve $\displaystyle \frac{dT}{dx} \:=\:0$

We have: .$\displaystyle \frac{x}{\sqrt{x^2+900}} \:=\:\frac{3}{5} \quad\Rightarrow\quad 5x\:=\:3\sqrt{x^2+900} $

Square both sides: .$\displaystyle 25x^2 \:=\:9(x^2 + 900)\quad\Rightarrow\quad 25x^2 \:=\:9x^2 + 8100$

. . $\displaystyle 16x^2 \:=\:8100\quad\Rightarrow\quad x^2 \:=\:\frac{2025}{4}\quad\Rightarrow\quad x \:=\:\frac{45}{2} \:=\:\boxed{22.5\text{ meters}}$

(ii) Substitute into the formula in part (a).

$\displaystyle T \;=\;\sqrt{22.5^2 + 900} + \frac{3}{5}(75-22.5) + 2 \;\;=\;\;\boxed{71\text{ seconds}}$

d) Find the time taken if the boy runs from A to D and then D to C. He would run: .$\displaystyle 30 + 75$ meters at $\displaystyle \frac{5}{3}$ m/sec

This will take him: .$\displaystyle \frac{105}{\frac{5}{3}} \:=\:\boxed{63\text{ seconds}}$

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Very funny!

If the boy's only concern is getting from $\displaystyle A$ to $\displaystyle C$ as fast as possible,

. . he should all the way (and skip the math).

However, if this some sport where __some__ swimming is required,

. . then our solution provides the shortest time.