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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: ch on May 18, 2013, 10:16:16 am

Title: Help please: Maximum and Minimum problems.
Post by: ch on May 18, 2013, 10:16:16 am: Hi I am having troubling with a maximum minimum question, just wondering if anyone could help the question is as follows:
Joseph is competing in a surf lifesaving competition race. Heis standing on a long,narrow point of landwhich juts 5 km out to sea directly opposite the surf club. The finish line is 8 kmalong the shore from
the surf club. If Joseph can swim at 7 km/h and run at 16 km/h, find the position along the shore that he should aim for in order to minimise the time taken to complete the race.
The answer is 2.43 km from the surf club.
If any one knows how to solve this could you please write back explaining your steps.Your help is greatly appreciated. :)
Title: Re: Help please: Maximum and Minimum problems.
Post by: Phy124 on May 18, 2013, 03:26:28 pm: Let's have a look at the diagram below.

(http://content.screencast.com/users/phy124/folders/Jing/media/81f2a0b4-f4bd-42b6-a8fe-57647c01710e/2013-05-18_1525.png)

The blue lines represent the swim leg and the red lines represent the running leg. (The green dot is the start and the orange the finish)

We are trying to find the point along the shore, which I've labeled $x$ , that will result in the minimum time taken.

This mean we need to find time as a function of $x$ .

The total distance that competitor will take is the summation of the swim leg and the running leg.

The distance for the swim leg is given by $\sqrt{25 + x^2}$ (derived in the above picture, using a right angled triangle of with know side lengths of $5$ and $x$ ) and the distance for the run leg is $8-x$

We are given the speeds that each leg can be done at and so we can find the time taken for each leg then add them.

The swim leg has a distance of $\sqrt{25 + x^2}$ and can be done at a speed of 7 km/h, therefore $t = \frac{d}{v} = \frac{\sqrt{25 + x^2}}{7}$

The running leg has a distance of $8-x$ and can be done at 16 km/h, therefore $t = \frac{d}{v} = \frac{8-x}{16}$

Hence the total time taken is $t(x) = \frac{\sqrt{25 + x^2}}{7} + \frac{8-x}{16}$

We want to find the minimum of this function i.e. the value of $x$ for which $t'(x) = 0$ , which will calculate to approximately 2.43km.

edit: removed an unfinished sentence that I accidentally left in there, awks :P
Title: Re: Help please: Maximum and Minimum problems.
Post by: ch on May 18, 2013, 07:05:58 pm: Thankyou so much for your help.