Chapter 4.6, Problem 35E

To determine

To calculate: Minimize the cost of pipeline. If an oil refinery is located on the north bank a straight river that is $2\text{km}$ wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river $6\text{km}$ east of the refinery. The cost of laying pipe is $\raisebox{1ex}{$\$400,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ over land to a point $P$ on the north bank $\raisebox{1ex}{$\$800,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ under the river to the bank.

Answer to Problem 35E

For $x=\frac{2}{\sqrt{3}}$ the cost of pipeline is minimum

Explanation of Solution

Given information:

An oil refinery is located on the north bank a straight river that is $2\text{km}$ wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river $6\text{km}$ east of the refinery. The cost of laying pipe is $\raisebox{1ex}{$\$400,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ over land to a point $P$ on the north bank $\raisebox{1ex}{$\$800,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ under the river to the bank.

Formula used:

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is:

  

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

Speed: the distance covered per unit time is called speed. Speed is directly proportional to distance and inversely proportional to time.

Speed =distance/time, Time =distance/speed, Distance =speed x time

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at $f^{\prime }(x)>0$ every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Calculation:

As per the given problem

Draw the diagram of an oil refinery is located on the north bank a straight river that is $2\text{km}$ wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river $6\text{km}$ east of the refinery. The cost of laying pipe is $\raisebox{1ex}{$\$400,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ over land to a point $P$ on the north bank $\raisebox{1ex}{$\$800,000$}\!\left/ \!\raisebox{-1ex}{$\text{km}$}\right.$ under the river to the bank.

  

Recall that,

Pythagorean Theorem: The sum of the squares on the legs of the right angled triangle is equal to the square on the side opposite to the right angle triangle. That is

  ${\left(H\right)}^2={\left(P\right)}^2+{\left(B\right)}^2$

in right angled triangle $△BAC$

  $\begin{array}{l}{\left(AB\right)}^2={\left(BC\right)}^2+{\left(AC\right)}^2\\ y^2=x^2+{\left(2\right)}^2\\ =x^2+4\end{array}$

Take square root on both sides,

  $y=\pm \sqrt{x^2+4}$

Because length can’t be negative,

Therefore,

  $y=\sqrt{x^2+4}$

The cost $C$ , which is hundreds of thousands of dollars is given by,

  $\begin{array}{l}C=\left(6-x\right)\times 4+y\times 8\\ =24-4x+8y\end{array}$

Substitute $t=0.36$

  $y=\sqrt{x^2+4}$ and simplified

  $C\left(x\right)=24-4x+8\sqrt{x^2+4}\mathrm{......}\left(1\right)$

Let $f$ be a differentiable function defined on an interval $I$ and let $a\in I$ .

Then

1. $x=a$ is a point of local maximum value of $f$, if
1. $f^{\prime }(a)=0$ and
2. $f^{\prime }(x)$ changes sign from positive to negative as $x$ increases through $a$ , i.e. if $f^{\prime }(x)>0$ at every point sufficiently close to and to the left of $a$ , and $f^{\prime }(x)<0$ at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local maxima
2. $x=a$ is a point of local maximum value of $f$, if $f^{\prime }(a)=0$ and
1. $f^{\prime }(x)$ changes sign from negative to positive as $x$ increases through $a$ , i.e. if $f^{\prime }(x)<0$ at every point sufficiently close to and to the left of $a$ , and at every point sufficiently close to and to the right of $a$ , then $a$ is a point of local minima.
3. $f^{\prime }(a)=0$ and If $f^{\prime }(x)$ does not change sign as $x$ increases through $a$ , then $a$ is neither a point of local maxima nor a point of local minima.
4. • If $f^{″}(a)>0$ then $f$ has a local minimum at $x=a$
   • If $f^{″}(a)<0$ then $f$ has a local maximum at $x=a$

Differentiate on both sides,

  $\begin{array}{l}{C}^{\prime }\left(x\right)=-4+\frac{8\times 2x}{2\sqrt{x^2+4}}\\ =-4+\frac{8x}{\sqrt{x^2+4}}\mathrm{......}\left(2\right)\end{array}$

Solve for $C^{\prime }\left(x\right)=0$ , and simplified

  $\begin{array}{l}-4+\frac{8x}{\sqrt{x^2+4}}=0\\ \Rightarrow \frac{8x}{\sqrt{x^2+4}}=4\\ \Rightarrow \sqrt{x^2+4}=2x\end{array}$

Square both sides,

  $\begin{array}{l}-4+\frac{8x}{\sqrt{x^2+4}}=0\\ \Rightarrow \frac{8x}{\sqrt{x^2+4}}=4\\ \Rightarrow \sqrt{x^2+4}=2x\\ \Rightarrow x^2+4=4x^2\\ \Rightarrow 3x^2=4\\ \Rightarrow x^2=\frac{4}{3}\end{array}$

Take square root on both sides,

  $x=\pm \frac{2}{\sqrt{3}}$

Because length can’t be negative,

Therefore,

  $x=\frac{2}{\sqrt{3}}$

Differentiate equation $\left(2\right)$ with respect to $t$

  $\begin{array}{l}{C}^{″}\left(x\right)=0+\frac{\sqrt{x^2+4}\times 8-8x\times \frac{x}{\sqrt{x^2+4}}}{{\left(\sqrt{x^2+4}\right)}^2}\\ =\frac{\left(x^2+4\right)8-8x^2}{\sqrt{x^2+4}{\left(\sqrt{x^2+4}\right)}^2}\\ =\frac{32}{\sqrt{x^2+4}{\left(\sqrt{x^2+4}\right)}^2}\end{array}$

Therefore,

  $C^{″}\left(x\right)=\frac{32}{\sqrt{x^2+4}{\left(\sqrt{x^2+4}\right)}^2}\ge 0$

For $x=\frac{2}{\sqrt{3}}$ the cost of pipeline is minimum

Substitute $x=\frac{2}{\sqrt{3}}$ in equation $\left(1\right)$ and simplified

  $\begin{array}{l}C\left(x\right)=24-4\times \frac{2}{\sqrt{3}}+8\sqrt{{\left(\frac{2}{\sqrt{3}}\right)}^2+4}\\ =24-\frac{4}{\sqrt{3}}\\ \approx \$21.61\end{array}$

Conclusion:

Thus for $x=\frac{2}{\sqrt{3}}$ the cost of pipeline is minimum