Very tricky minimization/maximization question. I need u

Q

Queenisabella87

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Nov 6, 2010
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A ship is anchored 4 miles off of straight shore. Down the shore 9 miles another ship is anchored 8 miles out to sea.

A landing craft leaves the first boat drops a person on the shore and continues to the second ship.



How far down shore should the person dropped to minimize he distance the landing craft travels??


As far as my work here, I figured out I may have to use a right triangle in the begining.

but I'm stuck, any body who knows how to begin this and can direct me to the answer or step by step solutions please help.

For what its worth I tried this all day and couldn't get a start on it.

:|
 
You have two right triangles.

The distance from the first boat to shore is \(\displaystyle \sqrt{x^{2}+16}\)

The distance from this point back to the second boat is \(\displaystyle \sqrt{(9-x)^{2}+64}\)

This is what must be minimized.

\(\displaystyle D(x)=\sqrt{x^{2}+16}+\sqrt{(9-x)^{2}+64}\)

\(\displaystyle D'(x)=\frac{x-9}{\sqrt{x^{2}-18x+145}}+\frac{x}{\sqrt{x^{2}+16}}\)

\(\displaystyle =\frac{x\sqrt{x^{2}-18x+145}+(x-9)\sqrt{x^{2}+16}}{\sqrt{(x^{2}+16)(x^{2}-18x+145)}}\)

Set the numerator to 0 and solve for x. It looks worse than it is.

Square both sides and get:

\(\displaystyle x^{2}(x^{2}-18x+145)=(x-9)^{2}(x^{2}+16)\)

Now, finish solving for x.

It whittles down to a quadratic you can factor.
 
Okay so from your last part, when I solved for x

I got the following ::

x[sup:1xohv67h]2[/sup:1xohv67h] (x[sup:1xohv67h]2[/sup:1xohv67h]-18x+43) + 18x - 97 = 0


now I set all 3 parts to 0

and i got 2 x's

x = 5.3 feet

and the other x is from the quadratic x[sup:1xohv67h]2[/sup:1xohv67h]-18x+43


is that right?
 
No. Sorry. :(

The resulting quadratic is \(\displaystyle 48x^{2}+288x-1296\)

\(\displaystyle 48(x-3)(x+9)=0\)
 
Hello, Queenisabella87!

There is a marvelous back-door approach that is not allowed in a Calculus course.

A ship is anchored 4 miles off of straight shore.
Down the shore 9 miles another ship is anchored 8 miles out to sea.
A landing craft leaves the first boat drops a person on the shore and continues to the second ship.

How far down shore should the person be dropped to minimize the distance the landing craft travels?
Click to expand...
Code: 
                            * B
                          * |
                        *   |
    A *               *     | 8 
      | *           *       |
    4 |   *       *         |
      |     *   *           |
    C * - - - * - - - - - - * D 
          x   P *    9-x    |
                  *         |
                    *       |
                      *     | 8
                        *   |
                          * |
                            * B

The first ship is at \(\displaystyle A\); the second is at \(\displaystyle B\).
The landing craft at \(\displaystyle A\) goes to point \(\displaystyle P\) on the shore and then to ship \(\displaystyle B\).
We want to minimize the distance \(\displaystyle AP + PB\).


Here is the Without-Tears method.

Reflect point \(\displaystyle B\) across the shoreline \(\displaystyle CD\) to point \(\displaystyle B'.\)
Draw line \(\displaystyle AB'\) ntersecting \(\displaystyle CD\) at \(\displaystyle P.\)
. . \(\displaystyle P\) is the desired point.

\(\displaystyle \text{We have: }\:CD = 9.\)

\(\displaystyle \text{Let }x = CP,\;9-x = PD.\)
\(\displaystyle \text{From similar right triangles: }\:\frac{x}{4} \:=\:\frac{9-x}{8} \quad\Rightarrow\quad x \:=\:3\)

 
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