minimum value

**bobby77** · November 5th 2006, 09:58 PM

need help...

**earboth** · November 5th 2006, 10:31 PM

Hello, bobby,

I label the island I.

The total distance is d = IC + CD

The total energy is e = 2*IC + CD

Let BC = x

Then you can calculate the energy by:

$e(x) = 2 \cdot \sqrt{5^2+x^2} + (13-x)$

You'll get the minimum if e'(x) = 0. So you need first the derivative of e (use chain rule!):

$e'(x) = 2 \cdot \frac{1}{2}\cdot \left( 5^2+x^2 \right)^{-\frac{1}{2}} \cdot 2x -1$

Now e'(x) = 0:

$2 \cdot \frac{1}{2}\cdot \left( 5^2+x^2 \right)^{-\frac{1}{2}} \cdot 2x -1=0$
$\frac{2x}{\sqrt{25+x^2}} =1$ . Multiply both sides by the denominator and afterwards square both sides:
$4x^2 = 25+x^2$ . Solve for x. You'll get 2 values for x. The negative one is senseless in the described situation.

I got $x=\frac{5}{3}\cdot \sqrt{3}\approx 2.887 \text{ km}$

EB

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