Laying cable across a river

Question

This is a very interesting calculus word problem that I came across in an old textbook of mine. So I know its got something to do with minimising distance, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. However, I did manage to make a picture or diagram of it.

Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

A power house, $P$, is on one bank of a straight river $200\textrm{ m}$ wide, and a factory, $F$, is on the opposite bank $400\textrm{ m}$ downstream from $P$.

The cable has to be taken across the river, under water at a cost of $\text{\$}6/\textrm{m}$.

On land the cost is $\text{\$}3/\textrm{m}$.

What path should be chosen so that the cost is minimized?

Edit: Thanks guys, I think I found out the answer. I shall post it on MSE.

Answer 1

So here is the diagram:

Please note the difference between P and P'

Let's look at the extreme cases first:

Straight across the river is $ \$200 m \cdot \$ 6m = \$1200$

$+ 400 m \ \text{on land} \cdot \$ 3m = \$1200$

Total: $ \$ 2400$

If we go diagonal, then by Pythagoras, it's $200 \sqrt 5 m \cdot \$6m = 1200 \sqrt 5 \approx \$2683$

Let the point straight across from P be P'.

Let's choose a landfall point Q, somewhere P' and F.

Let distance from P' to Q be q.

So total distance is

$\sqrt{200² + q²}$ (diagonal under the river) + 400 - q (on land)

Cost is $6 \cdot \sqrt{200² + q²} + 3 \cdot (400 - q)$

We want to minimize that.

First we can divide by 3 -- same value for q will be maximum.

$2 \cdot \sqrt{200² + q²} + 400 - q$

Derivative is

$2q \cdot \sqrt{200² + q²} - 1$

We set that to 0

$2q \cdot \sqrt{40000+q²} = 1$

$2q = \sqrt{40000+q²}$

Square:

$4 q² = 40000 + q^2$

$3 q² = 40000$

$q = 200 \sqrt 3$

Length under water is then $400 \sqrt 3 \approx 231$

and length on land is $400 - 200 \sqrt 3 \approx 284$

Cost is $6 \cdot 400\sqrt 3 + 3 (400 - 200\sqrt 3) = \$2239.23$

The triangle $P-P'-Q$ has sides $200 - 200\sqrt 3 - 400 \sqrt 3$

and therefore it is a 30-60-90 triangle.

The angle QPQ' = 30°.

Answer 2

Let me just get you started on the problem.

Let $x$ be the length of the cable on land, and $y$ be the length of the underwater cable.

Then, the cost, obviously, is $$C(x,y)=3x+6y$$

Can we find any other connection between $x$ and $y$? Of course we can! We know that $y$ is the hypothenuse of a right angled triangle. One of its sides is the width of the river, and the other is $400 - x$, meaning that $$y^2=200^2 + (400-x)^2$$

From these two equations (and the fact that $y>0$), you can determine the cost of the cable as a function of $x$, i.e. you get $C(x)$ which is a nice function of $x$. Then, all you have to do is find its minimum.

Answer 3

You want to minimise the sum of the line across the water and the line along the river bank, each weighted by the unit cost for each section. Assuming the cable can only be run in a straight line across the river, your objective function would be:

$$Cost=6\times\sqrt{200^2+q^2}+3\times(400-q)$$ To minimise, take the first derivative with respect to q and let it equal zero. Solve for q. Note the constraint that $q>=0$. Check for a minimum turning point using the second derivative with respect to q.

Answer 4

The length between P and Q is $\sqrt{q^2+200^2}$ (pythagorean theorem)

And the distance between F and Q is $400-q$

Therefore the cost function is $C(q)=6 \cdot \sqrt{q^2+200^2}+3\cdot (400-q)$

Take the derivative and set it equal to zero. Then solve for q.

Answer 5

We can borrow the optical refraction formula

Here the cost of cable can be considered to be the cost of time for light to go through water. Light always takes the fastest route.

let $\theta$ be the angel between cable and the banks, then we must have $\cos{\theta}=\frac{3}{6}*\cos{0}=\frac{1}{2}$. So $\theta=\frac{\pi}{6}$.

Answer 6

Let $\theta$ be angle $P'PQ$. Then $PQ = 200\sec\theta$, and $FQ = 400 - 200\tan\theta$. We want to minimise the cost, which is $$C(\theta)=6PQ+3FQ = 1200\sec\theta+1200-600\tan\theta$$ The derivative is $$\frac{dC}{d\theta} = 1200\sec\theta\tan\theta - 600\sec^2\theta$$ and setting this to zero gives $2\tan\theta=\sec\theta$, or $\sin\theta=\frac12$.