# Optimization Problem

• Nov 9th 2012, 06:22 AM
Alpharius120
Optimization Problem
Hi, first time poster.

So I've been wrestling with this problem for almost a week. It shouldn't be too difficult, there's just this one little aspect that throws me off. I can't tell if I can't figure out how to set it up, or if I'm just solving for the wrong variable.

Joel has fallen off of his “Little Mermaid” floaty 200 feet from shore in Lake Travis. He cannot swim. Cheryl is at a point 200 feet down the shore from the point closest to Joel. She can run 18 ft/s and can swim at a rate of 5 ft/s.

To what point on the shore should you run before diving into the lake if you want to reach Joel as quick as possible?

• Nov 9th 2012, 06:43 AM
richard1234
Re: Optimization Problem
Suppose Cheryl runs a distance $D_{run} = x$. Then we can compute the distance she swims in terms of x by building a right triangle with sides 200, 200-x, and D_swim. Hence,

$D_{swim} = \sqrt{200^2 + (200-x)^2}$

The amount of time taken is

$T = \frac{x}{18} + \frac{\sqrt{200^2 + (200-x)^2}}{5}$ (units are in seconds).

Simplify the expression for T, find T'(x), find critical points.
• Nov 9th 2012, 07:14 AM
Alpharius120
Re: Optimization Problem
When I took the derivative of T I got:
$T'(x) = (x-200)/(5*sqrt(x^2- 400x + 80000))+(1/18)$
but there are no critical points of this function.

Did I differentiate wrong or am I misunderstanding what you're saying I have to do?

Thanks for the response though!
• Nov 9th 2012, 07:28 PM
richard1234
Re: Optimization Problem
Your derivative is correct. Turns out there is a local minimum, so you may want to solve T'(x) = 0 again. I would probably use a calculator or computer.

Btw, according to WolframAlpha, the global minimum occurs at $x = 200 - \frac{1000}{\sqrt{299}}$.
• Nov 9th 2012, 07:54 PM
MarkFL
Re: Optimization Problem
Given that:

$T'(x)=\frac{x-200}{5\sqrt{200^2+(200-x)^2}}+\frac{1}{18}=0$

We may write:

$T'(x)=\frac{200-x}{\sqrt{200^2+(200-x)^2}}=\frac{5}{18}$

$\sqrt{200^2+(200-x)^2}=\frac{18(200-x)}{5}$

$200^2+(200-x)^2=\frac{324}{25}(200-x)^2$

$(200-x)^2=\frac{1000^2}{299}$

Take the positive root:

$200-x=\frac{1000}{\sqrt{299}}$

$x=200-\frac{1000}{\sqrt{299}}$