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 April 27th, 2013, 03:48 PM #1 Newbie   Joined: Apr 2012 Posts: 21 Thanks: 0 Optimisation and Extreme Value Theorem Hi all! I apologise if this is in the wrong forum, but I was unsure where else it would go. I'm struggling with this exam revision question that's loosely based around optimisation. The question is: Two towns are located near the straight of a shore of a lake. Their nearest distances to points on the shore are 1km and 2km respectively, and these points on the shore are 6km apart. A fishing pier is to be built on the shore of the lake and a straight road from each town to the pier is to be constructed. Where should the pier be positioned to minimise the sum of distances form the towns to the pier? Now the teacher gave us a hint and said to use the Extreme Value Theorem, and try to find the crit points and thus, the maximum and minimums. However, I get extremely lost somewhere near the beginning of that I've tried multiple times, but I generally fail somewhat. If anyone out there has a solution, I'd be eternally grateful; this question has been bugging me for a week now. I've uploaded a picture of the diagram we're given. IMG_0621.JPG Last edited by skipjack; November 22nd, 2017 at 09:57 AM.
 April 27th, 2013, 04:39 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Optimisation and Extreme Value Theorem A function describing the sum of the distances of the two towns from the pier is $\displaystyle f(x)\,=\,\sqrt{1^2\,+\,x^2}\,+\,\sqrt{2^2\,+\,(6\,-\,x)^2}$ Differentiate, set $f\,^\prime(x)$ equal to zero and solve for $x$: $\displaystyle f\,^\prime(x)\,=\,\frac{x}{\sqrt{1\,+\,x^2}}\,+ \,\frac{x\,-\,6}{\sqrt{x^2\,-\,12x\,+\,40}}\,=\,0$ $\displaystyle x^2(x^2\,-\,12x\,+\,40)\,=\,(x^2\,-\,12x\,+\,36)(1\,+\,x^2)$ $\displaystyle \cancel{x^4}\,-\,\cancel{12x^3}\,+\,40x^2\,=\,x^2\,-\,12x\,+\,36\,+\,\cancel{x^4}\,-\,\cancel{12x^3}\,+\,36x^2$ $\displaystyle 3x^2\,+\,12x\,-\,36\,=\,0$ $\displaystyle x^2\,+\,4x\,-\,12\,=\,0$ $\displaystyle (x\,+\,6)(x\,-\,2)\,=\,0\,\Rightarrow\,x\,=\,2$ Last edited by skipjack; November 22nd, 2017 at 11:34 AM.
 April 27th, 2013, 04:54 PM #3 Newbie   Joined: Apr 2012 Posts: 21 Thanks: 0 Re: Optimisation and Extreme Value Theorem ... I'm speechless. That looks so easy. Thank you so much. I was doing double differentials, derivative tests, and everything in between. I appreciate it mate.
 April 27th, 2013, 05:10 PM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond Re: Optimisation and Extreme Value Theorem You may need to confirm that x = 2 is a minimum. You can use the second derivative test.
 April 28th, 2013, 01:34 AM #5 Newbie   Joined: Apr 2012 Posts: 21 Thanks: 0 Re: Optimisation and Extreme Value Theorem I will do exactly that. Thank you sir.
 November 21st, 2017, 05:09 PM #6 Newbie   Joined: Nov 2017 From: Kamloops, BC, Canada Posts: 1 Thanks: 0 Answer is cut off? Hallo, I'm doing the same question, but still confused due to how the answer from "greg" is "... from the pier is Differentiate, ....". Why is it cutoff and not showing the function? Last edited by skipjack; November 22nd, 2017 at 10:05 AM.
November 21st, 2017, 05:31 PM   #7
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Quote:
 Originally Posted by roydencooke Hallo, Im doing the same question but still confused due to how the answer from "greg" is "... from the pier is Differentiate, ....". Why is it cutoff and not showing the function?
updated old latex tags and some syntax for clarity

note that the post is from 2013 ...

Quote:
 Originally Posted by greg1313 A function describing the sum of the distances of the two towns from the pier is $f(x)\,=\,\sqrt{1^2\,+\,x^2}\,+\,\sqrt{2^2\,+\,(6-x)^2}$ Differentiate, set $f\,^\prime(x)$ equal to zero and solve for $x$: $\displaystyle f\,^\prime(x)\,=\,\frac{x}{\sqrt{1\,+\,x^2}}\,+ \,\frac{x\,-\,6}{\sqrt{x^2\,-\,12x\,+\,40}}\,=\,0$ $x^2(x^2\,-\,12x\,+\,40)\,=\,(x^2\,-\,12x\,+\,36)(1\,+\,x^2)$ $\cancel{x^4}\,-\,\cancel{12x^3}\,+\,40x^2\,=\,x^2\,-\,12x\,+\,36\,+\,\cancel{x^4}\,-\,\cancel{12x^3}\,+\,36x^2$ $3x^2\,+\,12x\,-\,36\,=\,0$ $x^2\,+\,4x\,-\,12\,=\,0$ $(x\,+\,6)(x\,-\,2)\,=\,0\,\Rightarrow\,x\,=\,2$

Last edited by skipjack; November 22nd, 2017 at 11:38 AM.

 November 22nd, 2017, 11:20 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 With a diagram that shows where the pier should be (see below), it's easy to see that tan($\theta$) = 2, so that $x$ = 2. Roads.JPG The lower right corner in the diagram is the reflection of T$_2$ in the shore of the lake, and so has a fixed position. The pier needs to lie on the straight line between T$_1$ and that position.

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