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A pipeline is to be laid from a point $A$ on one bank of a river of width $1$ unit to a point $B$ $2$ units downstream on the opposite bank, as shown in Figure 8.40.

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Because it costs more to lay the pipe under water than on dry land, it is proposed to take it in a straight line across the river to a point $C$ and then along the river bank to $B$. If it costs $\alpha ~\%$ more to lay a given length of pipe under the river than along the bank, write down a formula for the cost of the pipeline, specifying the domain of the function carefully. What recommendation would you make about the position of $C$ when (a) $\alpha = 25$, (b) $\alpha = 10$?

I may have the wrong working equation:

Cost = length of pipe x cost/length

assuming cost/length is $1/unit, therefore,

Cost(underwater) = Cost(riverbank) + (0.25 x Cost(riverbank))

$$\frac{dL}{dx}=\frac{d}{dx} \left( \sqrt{x^2 + 1} + (2-x) \right) = \frac{x}{\sqrt{x^2 + 1}} - 1$$

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  • $\begingroup$ you are on the right track, you just forgot to multiply by $\alpha$: $\frac{dL}{dx}=\frac{(1+\alpha) x}{100\sqrt{x^2+1}}-1$ $\endgroup$ – Vasya May 15 '18 at 12:51
  • $\begingroup$ @Vasya $\frac{\alpha}{100}$. $\endgroup$ – Rodrigo de Azevedo May 16 '18 at 19:27
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Let us think a bit more on this problem: Let's say that the pipe under water costs $\mathscr{K}_1$ per meter with a length of $\text{L}_1$ meter, the width of the river is $\mathscr{W}$. The costs for the pipe on land is $\mathscr{K}_2$ per meter with a length of $\text{L}_1$ meter.

So, we can write:

$$\mathscr{K}_{\space\text{total}}=\mathscr{K}_1\cdot\text{L}_1+\mathscr{K}_2\cdot\text{L}_2\tag1$$

For $\text{L}_1$, we can write:

$$\text{L}_1=\left|\text{AC}\right|=\sqrt{\mathscr{W}^2+x^2}\tag2$$

For $\text{L}_2$, we can write:

$$\text{L}_2=\left|\text{BC}\right|=2\cdot\mathscr{W}-x\tag3$$

So, we can rewrite $\left(1\right)$ as follows:

$$\mathscr{K}_{\space\text{total}}=\mathscr{K}_1\cdot\sqrt{\mathscr{W}^2+x^2}+\mathscr{K}_2\cdot\left(2\cdot\mathscr{W}-x\right)\tag4$$

Let's find:

$$\frac{\partial\mathscr{K}_{\space\text{total}}}{\partial x}=0\space\implies\space x=\frac{\sqrt{5\cdot\left(\mathscr{K}_1\cdot\mathscr{K}_2\cdot\mathscr{W}\right)^2-\left(\mathscr{K}_1^2\cdot\mathscr{W}\right)^2}-2\cdot\mathscr{K}_2\cdot\mathscr{W}}{\mathscr{K}_1^2-\mathscr{K}_2^2}\tag5$$

Now, we also know that:

$$\mathscr{K}_1=\frac{100+\left(\alpha\space\text{%}\right)}{100}\cdot\mathscr{K}_2\tag6$$

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