# math question help (1 Viewer)

#### poptarts12345

##### Member
1. The point A lies on the positive half of the x-axis, and the point B lies on the positive half of the y-axis, and the interval AB passes through the point P(5,3). Find the coordinates of A and B so that triangle AOB has minimum area.

2. A man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is a further 20km down from the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land.

#### ultra908

##### Active Member
1. Let A be a variable point on the x axis, (x,0). Notice for any A, there can only be one B for which AB passes through (5,3). Thus we don't really have to worry about B for now. Then find an equation for area of triangle AOB, and its min using calculus. Using ur min value of x, find A and B!

2. Draw a diagram (assuming the coast is a straight line). Let x be the distance from A he lands. Find an equation for the time of the trip wrt to x,. Then find its min using calculus and thus the distance x.

#### Hannah0605

##### New Member
Send your question on Airstudy and see whether someone can help. They have a request teacher's function.

#### CM_Tutor

##### Well-Known Member
1. The point A lies on the positive half of the x-axis, and the point B lies on the positive half of the y-axis, and the interval AB passes through the point P(5,3). Find the coordinates of A and B so that triangle AOB has minimum area.
Let A be at $\bg_white (a, 0)$ and B be at $\bg_white (0, b)$ where a and b are positive. Since the line AB passes through $\bg_white P(5, 3)$, we actually know that $\bg_white a > 5$ and $\bg_white b > 3$.
(We know this as (i) if $\bg_white a = 5$ then AP is vertical and so does not pass through B and (i) if $\bg_white a < 5$ then AP has a positive slope and so intersects the negative half of the y-axis, violating the restriction on the position of B. The same reasoning applies to restricting that $\bg_white b > 3$.)

Now, the line AB has equation $\bg_white \frac{x}{a} + \frac{y}{b} = 1$ as the coordinates for A and B satisfy this linear equation. Further, the triangle AOB is right-angled at O and so has area given by $\bg_white A = \frac{ab}{2}$. So, we need only re-write this expression in two variables and apply calculus or other methods to find the required minimum. Since P lies on AB, we know that:

\bg_white \begin{align*} \frac{5}{a} + \frac{3}{b} &= 1 \\ 5b + 3a &= ab \\ 3a &= ab - 5b = b(a - 5) \\ b&= \frac{3a}{a - 5} \\ \\ \text{So,} \quad A &= \frac{ab}{2} = \frac{3a^2}{2(a - 5)} \\ \frac{dA}{da} &= \frac{2(a - 5) \times 6a - 3a^2 \times 2}{4(a - 5)^2} \\ &= \frac{12a^2 - 60a - 6a^2}{4(a - 5)^2} \\ &= \frac{6a(a - 10)}{4(a - 5)^2} \\ &= \frac{3a(a - 10)}{2(a - 5)^2} \end{align*}

So, we have stationary points at $\bg_white a = 0$ and $\bg_white a = 10$ and the derivative is undefined at $\bg_white a = 5$. Fortunately, we know the domain for valid answers is $\bg_white a > 5$ and so we need only investigate $\bg_white a = 10$.

When a is near to but below 10, $\bg_white \frac{dA}{da} = \frac{3a(a - 10)}{2(a - 5)^2} = \frac{(+)(-)}{(+)(+)^2} < 0$
And, when a is near to but above 10, $\bg_white \frac{dA}{da} = \frac{3a(a - 10)}{2(a - 5)^2} = \frac{(+)(+)}{(+)(+)^2} > 0$

It follows that in the vicinity of $\bg_white a = 10$, A is first decreasing, then stationary, then increasing, meaning that the stationary point there is a local minimum, where $\bg_white b = \frac{3a}{a - 5} = \frac{3 \times 10}{10 - 5} = 6$ and $\bg_white A = \frac{3a^2}{2(a - 5)} = \frac{3 \times 10^2}{2(10 - 5)} = \frac{3 \times 100}{10} = 30$.

Thus, the minimum area triangle AOB has A at (10, 0), B at (0, 6), and has area of 30 square units.

If you have TEX with tikzpicture installed, here is the picture I would draw to accompany this solution:
\begin{tikzpicture}
\fill[blue!20] (0, 0) -- (8, 0) -- (0, 8);
\draw[latex-latex] (-1.5, 0) -- (10, 0) node {$x$};
\draw[latex-latex] (0, -1.5) -- (0, 9) node
{$y$};
\draw[thick,blue] (8, 0) -- (0, 8);
\node at (4.5, 5) {$\color{blue} \cfrac{x}{a} + \cfrac{y}{b} = 1 \color{black}$};
\node at (-0.3, -0.3) {$O$};
\node at (8.5, 0.3) {$A(a, 0)$};
\node at (0.8, 8) {$B(0, b)$};
\node at (5.8, 3) {$P(5, 3)$};
\fill[black] (5, 3) circle (0.1);
\fill[black] (8, 0) circle (0.075);
\fill[black] (0, 8) circle (0.075);
\draw[dashed] (0, 3) -- (5, 3);
\draw[dashed] (5, 0) -- (5, 3);
\draw[dotted] (-1, 8) -- (0, 8);
\draw[dotted] (8, -1) -- (8, 0);
\draw[<-] (0, -0.75) -- (3.7, -0.75);
\draw[<-] (8, -0.75) -- (4.3, -0.75);
\node at (4.05, -0.75) {$a$};
\draw[<-] (-0.75, 0) -- (-0.75, 3.7);
\draw[<-] (-0.75, 8) -- (-0.75, 4.3);
\node at (-0.75, 3.95) {$b$};
\node at (5, -0.3) {$5$};
\node at (-0.3, 3) {$3$};
\node at (2.5, 1.5) {Area$_{\Delta AOB} = \cfrac{ab}{2}$};
\end{tikzpicture}

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#### CM_Tutor

##### Well-Known Member
2. A man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is a further 20km down from the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land.
Firstly, let's look at the answer that is sought:

Let S be the starting point for the rowing boat, which is 6 km from point A on the shore, which is the closest point on the shore to S.. We know that B is 20 km from A along the shore, and let point P be located a distance of x km from A, so that the distance PB is (20 - x) km.

The attached file contains this information shown in a diagram. We seek the position of P (and thus the value of x) so as to minimise the travel time from S to B. Let T be the travel time (in hours) from S to B, made up of the travel time from S to P (tSP) and the travel time from P to B (tPB). In otherwise, we need to minmise:

$\bg_white T = t_{SP} + t_{PB}$

Defining the distances SP and PB (in km) as dSP and dPB, respectively, and remembering that distance equals speed times times, we can see that:

$\bg_white \text{Travelling } d_{PB} = 20 - x \text{ km at 10 km/h implies that } t_{PB} = \frac{d_{PB}}{10} = \frac{20 - x}{10}$

Now, applying Pythagoras' Theorem to triangle SAP, we can see that SA2 + AP2 = SP2:

\bg_white \begin{align*} 6^2 + x^2 &= {d_{SP}}^2 \\ d_{SP} &= \sqrt{x^2 + 36} \\ \text{So, } t_{SP} &= \frac{\sqrt{x^2 + 36}}{8} \\ \text{So, } T = t_{PB} + t_{SP} &= \frac{20 - x}{10} + \frac{\sqrt{x^2 + 36}}{8} \\ \frac{dT}{dx} &= \frac{0 - 1}{10} + \frac{1}{8} \times \frac{1}{2} (x^2 + 36)^{\frac{-1}{2}} \times (2x + 0) \\ &= \frac{-1}{10} + \frac{2x}{8 \times 2} \times \frac{1}{\sqrt{x^2 + 36}} \\ &= \frac{-1}{10} + \frac{x}{8\sqrt{x^2 + 36}} \end{align*}

To find stationary points, we need to set this derivative to zero:

$\bg_white \frac{-1}{10} + \frac{x}{8 \sqrt{x^2 + 36}} = 0$
$\bg_white \frac{x}{8\sqrt{x^2 + 36}} = \frac{1}{10}$
$\bg_white \frac{40x}{8} = \frac{40\sqrt{x^2 + 36}}{10}$
$\bg_white 5x &= 4\sqrt{x^2 + 36}$
$\bg_white 25x^2 &= 16(x^2 + 36)$
$\bg_white (25 - 16)x^2 = 16 \times 36 \implies x^2 = \frac{16 \times 4 \times 9}{9} = 64$
$\bg_white x = 8 \text{ as } x \geqslant 0$

So, there is a stationary point at (x = 8 km, T = 2 h 27 min)

Following the path SAB (corresponding to x = 0), the rower rows 6 km taking 45 min and then runs 20 km taking 2 h, so that T = 2 h 45 min.
Following path SB (rowing directly from S to B, corresponding to x = 20), the distance rowed is (62 + 202)0.5 = 20.8806... km taking a little over 2 h 36 min. It follows that the stationary point is a minimum, and thus the rower should row to point P, 8 km from A towards B.

Of course, that x = 8 corresponds to a minimum can also be established by the usual methods of examining the behaviour of dT/dx in the vicinity of x = 8 or by showing that the second derivative is positive.

Note, however, that we have assumed that the shore line from A to B is a straight line with AB perpendicular to AS. I believe that this is what is intended, but the question does not actually specify this and if the shoreline is different from this then the answer may change.

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