My Math Forum Determine the Minimum value of P

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 March 25th, 2017, 02:02 AM #1 Newbie   Joined: Mar 2017 From: Scotland Posts: 8 Thanks: 0 Determine the Minimum value of P Hello, Sorry if this is in the wrong section; I don't know the difference between calculus and precalculus I'm having difficulties with a question of mine: The cost Â£P million, of laying 1km of pipe for a water main is given by the formula P = 4000/9a + 4a where a is the cross-sectional area of the pipe in square metres. Use calculus to find the minimum value of P, justifying your answer. I understand how to differentiate it and making the derivative equal to 0 to get the maximum and minimum points but I'm struggling to solve for a and any further point after that. Thanks in advance, Any help is greatly appreciated.
 March 25th, 2017, 05:53 AM #2 Senior Member   Joined: Jun 2015 From: England Posts: 760 Thanks: 218 $\displaystyle P = \frac{{4000}}{9}{a^{ - 1}} + 4a$ $\displaystyle \frac{{dP}}{{da}} = - \frac{{4000}}{9}{a^{ - 2}} + 4 = 0$ Is what sort of equation which has how many roots? So how do you use calculus to tell whether the root is a max or a min? Thanks from Beams
March 26th, 2017, 02:28 AM   #3
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Quote:
 Originally Posted by studiot $\displaystyle P = \frac{{4000}}{9}{a^{ - 1}} + 4a$ $\displaystyle \frac{{dP}}{{da}} = - \frac{{4000}}{9}{a^{ - 2}} + 4 = 0$ Is what sort of equation which has how many roots? So how do you use calculus to tell whether the root is a max or a min?
I've gotten to that point and have reached:

a^-2 = 9/1000

I've no idea how to solve for a then finding the minimum value.

 March 26th, 2017, 03:36 AM #4 Senior Member   Joined: Jun 2015 From: England Posts: 760 Thanks: 218 $\displaystyle 4 - \frac{{4000}}{9}{a^{ - 2}} = 0$ multiply through by $\displaystyle {a^2}$ $\displaystyle 4{a^2} - \frac{{4000}}{9} = 0$ $\displaystyle {a^2} = \frac{{1000}}{9}$ This is a simple quadratic equation in a, can you not solve it for two values of a? What does the fact that dP/da = 0 mean? It means that the slope of the curve is zero. This happens at both a minimum and a maximum, but we don't know which. To find out we take the second derivative $\displaystyle If\quad \frac{{{d^2}P}}{{d{a^2}}}\quad is\;negative\;the\;po{\mathop{\rm int}} \;is\;a\;\max imum$ $\displaystyle If\quad \frac{{{d^2}P}}{{d{a^2}}}\quad is\;positive\;the\;po{\mathop{\rm int}} \;is\;a\;\min imum$ https://en.wikipedia.org/wiki/Derivative_test Thanks from Beams
 March 26th, 2017, 04:00 AM #5 Newbie   Joined: Mar 2017 From: Scotland Posts: 8 Thanks: 0 Alright, would this be correct? I've gotten a = (10âˆš10)/3 and a= -(10âˆš10)/3 I've found the second derivative to be: 2(4000/9)a^-3 When I sub in a I get 0.7589 which is greater than 0 which implies a maximum. Therefore P = Â£0.76 Million? Is that it? Also how do you use MathML? What I'm typing looks really ugly.
March 26th, 2017, 04:22 AM   #6
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Quote:
 Originally Posted by Beams Also how do you use MathML? What I'm typing looks really ugly.
If you quote a post, you can see the LaTeX syntax to use. Bit of a learning curve at first, but well worth it!

 March 26th, 2017, 04:26 AM #7 Math Team   Joined: Jul 2011 From: Texas Posts: 2,722 Thanks: 1376 Note that $a > 0$ in the given context of this problem, so you can ignore the negative solution. 2nd derivative > 0 indicates a minimum ... Thanks from Beams
March 26th, 2017, 04:34 AM   #8
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Quote:
 Originally Posted by skeeter 2nd derivative > 0 indicates a minimum ...
Ahh, I meant to write minimum, I have no idea what I was thinking when I typed that.

Quote:
 Originally Posted by Joppy If you quote a post, you can see the LaTeX syntax to use. Bit of a learning curve at first, but well worth it!
Thanks, I'll look for tutorials and try my best to learn how to use it.

Thanks for the help with this problem everyone, I understand it much better.

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