Calculus - Minimizing Distance

It can be seen that cost $y$ is a function of distance $q$ (as denoted in the diagram).

$y = 400-q+\sqrt{q^2+200^{2}}$

We can find the minimum of $y$ by taking its derivative and equating with zero:

$\frac{dy}{dq} = -1+\frac{2q}{\sqrt{q^2+200^{2}}} = 0$ .

Hence, $q = \frac{200 \sqrt{3}}{3}$ . The value of $q$ obtained is a minimum as as can be seen by the second derivative test.

Let $\theta$ be the angle from the power house. $\tan \theta = \frac{200 \sqrt{3}}{600} = \frac{\sqrt{3}}{3}$ . Thus, $\theta = \boxed{\pi/6}$ .

Just check the options.......The least cost is $2239.23.... :p

L= 400 ; W=200 ; c1= 3; c2 =6

Cost function Q= c1 (L- W tan p) + c2*W/(cos p)

making dQ/dp = 0

c1 W/(cos p)^2- c2 W* sin p /(cos p)^2= 0

Simplifying we get sin p = c1/c2

Plugging values c1=3 and c2=6 result sin p = 1/2; p = 30

It is interesting to note that angle to be chosen only depends on the cost.

Let's look at the extreme cases first:

Straight across the river is 200 m @ $6/m = $1200 plus 400 m on land @ $3/m = $1200, total $2400.

If we go diagonal, then by Pythagoras, it's 200 sqrt(5) m @ $6/m = 1200 sqrt(5) ~= $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 * sqrt(40000+q²) + 3 * (400 - q) We want to minimize that. First we can divide by 3 -- same value for q will be maximum. 2 * sqrt(40000+q²) + 400 - q Derivative is 2q / sqrt(40000+q²) - 1 We set that to 0 2q / sqrt(40000+q²) = 1 2q = sqrt(40000+q²) Square: 4 q² = 40000 + q² 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 * 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°.