# Math Help - Differential Calculus Problem

1. ## Differential Calculus Problem

This is a question out of a textbook and it would be greatly appreciated if you could point me in the right direction. Don't worry, I've done most of it - I just need a little guidance on what to do next.

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An oil platform is located at point O, at sea, 10km from nearest point P on a stretch of straight coastline. The oil refinery is at point R, 16km along the coast from point P (see diagram). A company needs to lay a pipeline from the platform to the refinery (ie. from point O to R). This is to consist of straight line sections. Laying pipeline underwater is 4/3 times as expensive as laying it on land:

(a) If the pipeline reaches the coast at point X (as shown in my diagram) a distance x km from P in direction of R, find an expression for the cost C (in dollars) of laying the pipeline in terms of x, given that laying the pipeline on land has a fixed cost of A dollars per kilometre. Here's what I've done so far (could you check if it's right?):

Using Pythagoras, OX = $\sqrt{100 + x^2}$

$C(x) = \frac{4}{3}\sqrt{100 + x^2} + (16 - x)$

(b) Find the cost of the least expensive route for laying the required pipeline. Here's what I've done:

$C(x) = \frac{4}{3}\sqrt{100 + x^2} + (16 - x)$

I differentiated it (hopefully I did it correctly):

$C'(x) = \frac{4x}{3\sqrt{100 + x^2}} - 1$

And simply solved for x (I used my calculator to do this):

$x = 11.339$

Next, I simply plugged this into my diagram to get the required dimensions:

$16 - 11.339 = 4.661$

So the required distance from X to R is 4.661 km.

$\sqrt{100 + 11.339^2} = 15.119$

So the required distance from O to X is 15.119 km.

$\frac{4}{3}(15.119) + 4.661 = 24.82$

Thanks.

2. Plug the value of x in the cost function you found at a to find the minimal costs

3. $C'(x) = \frac{4x}{3\sqrt{100 + x^2}} - 1$

"And simply solved for x (I used my calculator to do this)"
Why not solve it algebraically?

$\frac{4x}{3\sqrt{100 + x^2}} - 1=0$
$4x=3\sqrt{100 + x^2}$
$16x^2=9(100 + x^2)$
$7x^2=900$
$x^2=\frac{900}{7}$
$x=\frac{30}{\sqrt{7}}$, as x should be positive.

So your answer is $C(\frac{30}{\sqrt{7}})$

4. Yup, that's what I did and I got 24.819.

So the cost is about $24.82? Seems a little awkward (and a bit too cheap). 5. No, the answer is$24.819*A. Sorry I read over it, you need to have;
$C(x) = A(\frac{4}{3}\sqrt{100 + x^2} + (16 - x))$, so that you get $C'(x) = A(\frac{4x}{3\sqrt{100 + x^2}} - 1)$. The optimal value for x stays the same.