relative min prob: A starts 30 mi south of B, going north...

[DBraden7]

I have a problem I have to have worked by monday and I am completely stumped. I have figured out that I am going to have to use the distance formula and find the derivative then find the min. I can do the algebra part, i'm just having trouble setting it up. If anyone could give me the correct set up and explain why, I would be really appreciative.

At 1:00 P.M. ship A is 30 miles due south of ship B and is sailing north at a rate of 15 mi/hr. If ship B is sailing west at a rate of 10 mi/hr, find the time at which the distance d between the ships is minimal.

 

[soroban]

Re: Help with relative min problem please

Hello, DBraden7!

At 1:00 pm ship \(\displaystyle A\) is 30 miles due south of ship \(\displaystyle B\) and is sailing north at a rate of 15 mph.
If ship \(\displaystyle B\) is sailing west at a rate of 10 mph, find the time at which the distance \(\displaystyle d\)
between the ships is minimal.

First we need a good diagram . . .

        B   10t   Q
        * - - - - * 
          *       |
            *     | 30-15t
           d  *   |
                * |
                  * A 
                  |
                  | 15t
                  |
                  *
                  P

\(\displaystyle \text{Point }Q\text{ is directly north of point }P\)
. . \(\displaystyle \text{and }PQ = 30\text{ miles.}\)

\(\displaystyle \text{At 1:00, ship }A\text{ starts at point }P\text{ and heads north at 15 mph.}\)
\(\displaystyle \text{In }t\text{ hours, it has moved }15t\text{ miles to point }A.\)
. . \(\displaystyle \text{Then: }\;AQ \:=\:30-15t\)

\(\displaystyle \text{At 1:00, ship }B\text{ starts at point }Q\text{ and heads west at 10 mph.}\)
\(\displaystyle \text{In }t\text{ hours, it has moved }10t\text{ miles to point }B.\)
. . \(\displaystyle \text{Then: }\;BQ \:=\:10t\)

\(\displaystyle \text{Their distance, }d = AB\text{, is the hypotenuse of right triangle }BQA.\)

\(\displaystyle \text{Hence: }\;d \;=\;\sqrt{(10t)^2 + (30-15t)^2} \;=\;\left(325t^2 - 900t + 900\right)^{\frac{1}{2}}\)


\(\displaystyle \text{Differentiate and equate to zero: }\;d' \;=\;\tfrac{1}{2}\left(326t^2-900t+900\right)^{-\frac{1}{2}}(650t - 900) \;=\;0\)

\(\displaystyle \text{We have: }\;650t - 900 \:=\:0\quad\Rightarrow\quad 650t \:=\:900\quad\Rightarrow\quad t \:=\:\frac{18}{13}\)


\(\displaystyle \text{They will be the closest in }1\tfrac{5}{13}\text{ hours}\;\approx\;\text{1 hour, 23 minutes.}\)

\(\displaystyle \text{The time will be }\boxed{2:23\text{ pm}}\)

 

[DBraden7]

Thank you for the very good explanation. It helped me to understand the problem better.