Minimizing Costs (Word Problem)

**tivo1980** · December 8th 2006, 01:57 PM

This is a word problem that deals with (minimizing costs):

A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and $3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

HELP PLEASE!!!

**TriKri** · December 8th 2006, 03:11 PM

Originally Posted by tivo1980

This is a word problem that deals with (minimizing costs):

A power line is to be constructed from a power station at point A to an island at point C, which is 1 mi directly out in the water from a point B on the shore. Point B is 4 mi downshore from the power station at A. It costs $5000 per mile to lay the power line under water and $3000 per mile to lay the line under ground. At what point S downshore from A should the line come to the shore in order to minimize cost? Note that S could very well be B or A (Hint: The length of CS is sqrt(1 + x^2)

HELP PLEASE!!!

I guess that $ABC$ forms a right-angle triangle and that $S$ can be chosen anywhere along the line $AB$ .

Let's say $x$ is the distance in mi upshores from $B$ to $C$ .

The total cost $t\ =\ \sqrt{x^2 +1}\cdot \$ 5000 + (4-x)\cdot \$ 3000$ .

$t\ =\ (x^2 +1)^{0.5}\cdot \$ 5000 + (4-x)\cdot \$ 3000$ .

Minimize $t$ :

$\frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \$ 5000 - \$ 3000\ =\ 0$

$\frac{2 x}{2\sqrt{x^2+1}}\cdot \$ 5000\ =\ \$ 3000$

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$50 x^2\ =\ 36(x^2+1)$

$14 x^2\ =\ 36$

$x^2\ =\ \frac{18}{7}$

$x\ =\ 3\sqrt{\frac{2}{7}}$

Answer: $4-3\sqrt{\frac{2}{7}}$

**tivo1980** · December 8th 2006, 03:14 PM

Thank you Trikri.

**TriKri** · December 8th 2006, 03:23 PM

I must be really bored.

**tivo1980** · December 8th 2006, 03:54 PM

I have a question Trikri. What software did you use to present the math problem the way you did?

**earboth** · December 9th 2006, 12:34 AM

Originally Posted by tivo1980

I have a question Trikri. What software did you use to present the math problem the way you did?

Hello, Tivo,

have a look here: http://www.mathhelpforum.com/math-he...rial-latex.pdf

EB

**earboth** · December 9th 2006, 05:18 AM

Originally Posted by TriKri

I guess that $ABC$ forms a right-angle triangle and that $S$ can be chosen anywhere along the line $AB$ .

Let's say $x$ is the distance in mi upshores from $B$ to $C$ .

The total cost $t\ =\ \sqrt{x^2 +1}\cdot \$ 5000 + (4-x)\cdot \$ 3000$ .

$t\ =\ (x^2 +1)^{0.5}\cdot \$ 5000 + (4-x)\cdot \$ 3000$ .

Minimize $t$ :

$\frac{\partial t}{\partial x}\ =\ 0.5\cdot(x^2+1)^{-0.5}\cdot 2\cdot x\cdot \$ 5000 - \$ 3000\ =\ 0$

$\frac{2 x}{2\sqrt{x^2+1}}\cdot \$ 5000\ =\ \$ 3000$

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$50 x^2\ =\ 36(x^2+1)$

$14 x^2\ =\ 36$

$x^2\ =\ \frac{18}{7}$

$x\ =\ 3\sqrt{\frac{2}{7}}$

Answer: $4-3\sqrt{\frac{2}{7}}$

Hello, Trikri,

I don't agree with your result:

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$100 x^2\ =\ 36(x^2+1)$

$x^2\ =\ \frac{36}{64}$

$x\ =\frac{6}{8}$ . The negative result isn't very realistic with this problem.

Answer: $4-\frac{3}{4}=3.25\text{ miles}$ from A.

EB

**TriKri** · December 9th 2006, 07:26 AM

Originally Posted by earboth

Hello, Trikri,

I don't agree with your result:

$2 x\cdot 5\ =\ 2\sqrt{x^2+1}\cdot 3\ (\Rightarrow x \geq 0)$

$100 x^2\ =\ 36(x^2+1)$

Oh, I was a little tired last night ... I must go to bed earlier!

$x\ =\frac{6}{8}$ . The negative result isn't very realistic with this problem.

That's why I showed that $x \geq 0$

**tivo1980** · December 9th 2006, 09:15 AM

Thank you earboth for the document and the solution.

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