# Math Help - Optimization help

1. ## Optimization help

Im having some issues understanding an example in my book, mainly due to the fact that there is some missing explanation of what was done. The idea is I need to eventually use something similar to this to for another question to solve a minimum cost optimization problem.

A man launches his boat at point A on a straight river, 3km wide and wants to reach point B, 8km downstream. He can row at 6km/h and run at 8km/h

Rowing distance is $\sqrt{x^2 + 9}$
Running distance is $8 - x$

$time = \frac {distance}{rate}$

$\frac{\sqrt{x^2 + 9}}{6} + \frac{8 - x}{8}$

$T'(x) = \frac {x}{6\sqrt{x^2 + 9}} - \frac {1}{8}$

Find where x = 0
$T'(x) = 0 = \frac {x}{6\sqrt{x^2 + 9}} = \frac {1}{8}$

Im guessing in the example we want to land part to be on the other side which is why its moved over.

This is where I get confused, Im not sure what exactly has happened, but I think we have some how canceled the top of the left side fraction than halved both sides.

$4x = 3\sqrt{x^2 + 9}$

From here Im guessing we square both sides to remove the square root.
$16x^2 = 9(x^2 + 9)$

Than remove the square root from the right so subtract the left sided one
$7x^2 = 81$

The square root both sides
$x = \frac {9}{\sqrt{7}}$

2. Originally Posted by gk99
Im having some issues understanding an example in my book, mainly due to the fact that there is some missing explanation of what was done. The idea is I need to eventually use something similar to this to for another question to solve a minimum cost optimization problem.

A man launches his boat at point A on a straight river, 3km wide and wants to reach point B, 8km downstream. He can row at 6km/h and run at 8km/h

Rowing distance is $\sqrt{x^2 + 9}$
Running distance is $8 - x$

$time = \frac {distance}{rate}$

$\frac{\sqrt{x^2 + 9}}{6} + \frac{8 - x}{8}$

$T'(x) = \frac {x}{6\sqrt{x^2 + 9}} - \frac {1}{8}$

Find where x = 0
$T'(x) = 0 = \frac {x}{6\sqrt{x^2 + 9}} = \frac {1}{8}$

Im guessing in the example we want to land part to be on the other side which is why its moved over.

This is where I get confused, Im not sure what exactly has happened, but I think we have some how canceled the top of the left side fraction than halved both sides.

$4x = 3\sqrt{x^2 + 9}$

From here Im guessing we square both sides to remove the square root.
$16x^2 = 9(x^2 + 9)$

Than remove the square root from the right so subtract the left sided one
$7x^2 = 81$

The square root both sides
$x = \frac {9}{\sqrt{7}}$
$T'(x) = 0 = \frac {x}{6\sqrt{x^2 + 9}} = \frac {1}{8}$

At this stage it should be:
$T'(x) = 0 = \frac {x}{6\sqrt{x^2 + 9}} - \frac {1}{8}$
Note the last equal sign should be a minus sign.
That then gives:
$\frac {x}{6\sqrt{x^2 + 9}} = \frac {1}{8}$
Cross multiplying gives:
$8x = 6\sqrt{x^2 + 9}$
Then divide both sides by 2. Then continue as you did.

Cross multiplying:
If a/b = c/d then ad=bc

3. Thats what I thought, so its a typo in the book, as this example is from the book not something Ive done which is why Ive had a bit of trouble understanding how they got what they got.

4. I am having a small issue in a related problem except using cost.

An oil refinery is located on the north bank of a straight river that is 2km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. Cost of the laying the pipe is $400,000/km over land to point P and$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should P be located >

My equation to solve the cost is

$C(x) = 8\sqrt{4+ x^2} + 4(6-x)$

The derivative I find for this is

$C'(x) = \frac{8x}{\sqrt{4 + x^2}} + 4$

From here I get
$\frac{8x}{\sqrt{4 + x^2}} = 4$

Halve it.
$\frac{4x}{\sqrt{4 + x^2}} = 2$

$\frac{16x^2}{4 + x^2} = 4$

$16x^2 = 4(4 + x^2)$

$12x^2 = 16$

$\frac{16}{\sqrt{12}}$

This gives the answer approx 4.61 which is incorrect according to the answer book it should be approx 4.85 east of the refinery which means somewhere trying to solve it I mucked up somewhere but can't see where I did.

Also the other issue is I am using Maple which gives me a totally different result of $-\frac{2}{\sqrt{3}}$ which is also wrong.

5. Originally Posted by gk99
I am having a small issue in a related problem except using cost.

An oil refinery is located on the north bank of a straight river that is 2km wide. A pipeline is to be constructed from the refinery to storage tanks located on the south bank of the river 6km east of the refinery. Cost of the laying the pipe is $400,000/km over land to point P and$800,000/km under the river to the tanks. To minimize the cost of the pipeline, where should P be located >

My equation to solve the cost is

$C(x) = 8\sqrt{4+ x^2} + 4(6-x)$

The derivative I find for this is

$C'(x) = \frac{8x}{\sqrt{4 + x^2}} + 4$

From here I get
$\frac{8x}{\sqrt{4 + x^2}} = 4$

Halve it.
$\frac{4x}{\sqrt{4 + x^2}} = 2$

$\frac{16x^2}{4 + x^2} = 4$

$16x^2 = 4(4 + x^2)$

$12x^2 = 16$

$\frac{16}{\sqrt{12}}$ ... should be $\textcolor{red}{\frac{4}{\sqrt{12}} = \frac{2}{\sqrt{3}}}$
next time start a new problem w/ a new thread.