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October 30th, 2008, 02:40 AM   #1
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Minimum value

I am very green around the gills with calculus and i have this question that i have to find the number of (x) that's needs to be produced to minimize the cost of P

I am given

P(x) = 2x^3-5x^2+2

dy/dx = 6x^2-10x

0 = 6x^2-10x
10x = 6x^2
10 = 6x
x=10/6 = 5/3

P = 2*(5/3)^3 - 5*(5/3)^2 + 2
p = -2.6296

d2y/dx2 = 12x-10
=12*-2.6296-10
=-41.55

This is where i get stuck i have checked x and P on graphmatica and i can see where they are but not sure what to do with them now.
Can anyone help me ....
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October 30th, 2008, 05:49 AM   #2
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Re: Minimum value

Quote:
Originally Posted by ferret

P(x) = 2x^3-5x^2+2

dy/dx = 6x^2-10x

[color=#FF0000]CORRECT[/color]

0 = 6x^2-10x
10x = 6x^2
10 = 6x
x=10/6 = 5/3

[color=#FF0000]CORRECT

Now you are done. Why go further if not asked to? Be careful of not wasting time on one thing needed for another during an exam.[/color]

P = 2*(5/3)^3 - 5*(5/3)^2 + 2
p = -2.6296

[color=#FF0000]CORRECT[/color]

[color=#FF0000]WHY are you doing the rest? Don't just "do" calculus. Think about what you are doing and use it as the tool. Think about why you would do the second derivative and what purpose it might serve in this problem ..NONE. The question had only asked to find the value of x. You should have stopped there.[/color]

d2y/dx2 = 12x-10
=12*-2.6296-10
=-41.55

This is where i get stuck i have checked x and P on graphmatica and i can see where they are but not sure what to do with them now.
Can anyone help me ....
Dave is offline  
October 30th, 2008, 06:33 AM   #3
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Re: Minimum value

The second derivative check is used to see if the x obtained by solving f'(x)=0 is a minimum or a maximum (or none). It is necessary to check this since the problem specifies to find the minimum value of x, not the x where the derivation of P equals 0 - that is just a way of solving it.
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October 30th, 2008, 12:18 PM   #4
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Re: Minimum value

Quote:
Originally Posted by milin
The second derivative check is used to see if the x obtained by solving f'(x)=0 is a minimum or a maximum
Well, OK in general; however, this is not a complex function but a simple cubic [up to the right since positive leading coefficient, then down,then up general form]. The factors of the derivative yield two values for max/min, and the first is at zero and will have to be a max due to the general form, the second at 5/3 would have to be a min. That is, a max has to precede the min when clearly not a single point.
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October 30th, 2008, 01:56 PM   #5
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Re: Minimum value

Thanks for all your help i will go back and have a look at what you have said
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October 30th, 2008, 05:09 PM   #6
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Re: Minimum value

Quote:
Originally Posted by ferret
Thanks for all your help i will go back and have a look at what you have said
Just note what I said about factoring please. You have more than one value that satisfies 6x^2-10x = 0.
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